# 3. AC Fundamentals and Single-phase Circuits – Basic Electrical and Electronics Engineering

## AC Fundamentals and Single-phase Circuits

TOPICS DISCUSSED
• Concept of dc and ac
• Concept of frequency
• Time period
• Instantaneous value
• Average value and maximum value of an alternating quantity
• Sinusoidal and non-sinusoidal wave forms
• Concept of root mean square value
• Concept of phase and phase difference
• Single-phase ac circuits
• Series–parallel circuit containing resistance inductance and capacitance
• Concept of apparent power
• Real power and reactive power
• Resonance in ac circuits
• Solution of ac circuit problems
##### 3.1 AC FUNDAMENTALS

Now days electricity is generated in the form of ac (alternating current). The generated electricity is transmitted, distributed, and mostly utilized in the form of ac. In this chapter, the fundamental concepts of ac and ac circuits have been discussed.

#### 3.1.1 Introduction

We have known that current drawn from a battery is unidirectional. The polarities of the battery are marked +ve and –ve. When a particular load, say a lamp (represented by its resistance) is connected across the two terminals of the battery, current flows through the lamp in a particular direction. The magnitude of current as well as its direction remains constant with respect to time as long as the battery voltage remains constant. Such a current is known as steady or constant-value direct current.

Figure 3.1 Concept of dc and ac illustrated. (a) A battery connected to a resistive load; (b) direct current of constant magnitude; (c) direct current of variable magnitude; (d) a battery connected to a resistive load through a reversing switch; (e) alternating current of square–wave shape; (f) alternating current of rectangular wave shape

When the direction of current through a circuit continuously changes, such a current is called alternating current. The polarities of the ac supply source changes alternately and causes alternating current to flow through the load connected across the terminals. Fig. 3.1 (a) shows dc flowing through the load when connected across the battery terminals. Since the magnitude is assumed constant, it is represented through a graph as shown in Fig. 3.1 (b). Fig. 3.1 (c) shows the nature of the current when the switch is turned ‘ON’ and ‘OFF’ at regular constant intervals. By using a reverse switching arrangement, as shown in Fig. 3.1 (d), we can have an ac flowing through the load. Here, through a reversing switch, supply terminal 1 is connected to load terminal 2ʹ and supply terminal 2 is connected to load terminal 1ʹ. It can be noted that if the period of switching ON in both the directions in kept constant, the load current will be alternating in nature and its wave shape will be square or rectangular type as shown in Fig. 3.1 (e) and 3.1 (f).

To generate ac from an available dc source, we need an automatic switching arrangement. This is achieved by using electronic circuitry as in the case of inverters used as emergency lighting arrangement.

However, alternating current on a large scale is made available by using ac generators installed in power houses. AC generators are driven by turbines (gas, steam, water). Turbines are used to create a relative motion between a set of magnets and a set of coils. The rate of change of magnetic flux linkages or the rate of cutting of flux by the conductors of the coils causes EMF to be induced in the coil windings. The relative motion between the coils and the magnets producing a magnetic field can be created by making one system rotating with respect to the other. For example, we may have a stationary magnetic field system and inside the magnetic field we can place the coils which will be rotated by a prime mover (i.e., a turbine).

Alternately, the coils could be kept stationary and a set of magnets could be made rotating, thus causing EMF to be induced in the coils. We shall see the nature of EMF induced when we create a relative motion between a set of coils and a magnetic field. For simplicity we will consider only one coil rotating in a magnetic field created by a North and a South Pole.

#### 3.1.2 Generation of Alternating Voltage in an Elementary Generator

In Fig. 3.2 (a) is shown a coil having a few turns rotated in a magnetic field. If ϕ is the flux produced in Webers in the magnetic field and N is the number of turns of the coil, the flux linkage by the coil, i.e., the amount of flux passing through the coil will be (Nϕ) Webers. When the coil rotates, there is a change in the flux linkage. The induced EMF ‘e’ is the rate of change of flux linkage which can be expressed as

e = –d/dt (Nϕ) = –N dϕ/dt V

Figure 3.2 EMF induced in a coil rotated in a magnetic field. (a) Maximum flux linkage but minimum rate of change of flux linkage; (b) same as in (a); (c) the coil has rotated by an angle θ from its vertical position increasing the rate of change of flux linkage

Figure 3.3 Sinusoidal EMF induced in a coil rotating in a uniform magnetic field. (a) Coil rotating; (b) wave shape of the induced EMF

It may be seen from Fig. 3.2 (a) and 3.2 (b) that flux ϕ is perpendicular to the coil. When the coil rotates through an angle, say θ, from its vertical axis, as shown in Fig. 3.2 (c), the component of flux ϕ which then becomes perpendicular to the plane of the coil is ϕm cos θ. If θ is taken as ωt, ϕ = ϕm cos ωt.

In position aʹ bʹ of the coil ab, PQ is the component of flux ϕm, i.e., RQ that will link the coil. From Fig. 3.3 (a),

PQ = RQ cos θ = ϕm cos θ = ϕm cos ωt

Induced EMF,

e = –Ndϕ/dt = –Nd/dt(ϕm cos ωt)

= Nωϕm sinωt

where,                                    Em = Nω ϕm = N 2π f ϕm V

It is seen from eq. (3.1) that the induced EMF is sinusoidal in nature when the coil rotates in a uniform magnetic field as has been shown in Fig. 3.3 (b). For the initial position of rotation, i.e., when the coil plane is vertical to the direction of the flux, the EMF induced is minimum because a little change in angle θ does not cause much change in the flux linkage, or cutting of flux by the conductor is minimum. In the horizontal position of the coil, any small change in the coil angle causes a large change in the flux linkage or the cutting of flux by the conductor is maximum, and hence the induced EMF is the highest at that position.

#### 3.1.3 Concept of Frequency, Cycle, Time Period, Instantaneous Value, Average Value, and Maximum Value

One set of positive values and the subsequent one set of negative values of an alternating quantity constitute a cycle. The time taken for the generation of one cycle of EMF or flow of current caused due to such an EMF is called the time period, T. The total number of cycles of EMF or current produced per second is called the frequency, f. The relationship between time period, T and frequency, f can be found out as follows:

In T seconds the cycle produced is 1

In 1 second the cycle produced is 1/T

Since f is the cycle produced per second,

The value of an alternating quantity at any instant of time is called its instantaneous value. Such values are expressed in small lettering like e, i, etc. For sinusoidal waves, we may write

e = Em sin θ

and

i = Im sin θ

at

θ = 0°, e = Em sin 0°=0

at

θ = 90°, e = Em sin 90°=Em

Em is called the maximum value which occurs at θ = 90°, i.e., when the plane of the rotating coil is parallel to the magnetic field.

#### 3.1.4 Sinusoidal and Non-sinusoidal Wave Forms

We have seen earlier that when a coil rotates in a uniform magnetic field the EMF induced in the coil is sinusoidal in nature. The wave shape of an alternating voltage or current produced in an ac generator having uniform flux distribution is also sinusoidal in nature. However, an alternating quantity may be nonsinusoidal also. Any non-sinusoidal wave can be seen as consisting of a number of sinusoidal waves of different frequencies. Such component sine waves of a non-sinusoidal wave are called harmonic waves. In Fig. 3.4 (a) have been shown non-sinusoidal waves. In Fig. 3.4 (b) have been shown non-sinusoidal waves and their corresponding component sine waves. The component sine wave having the same frequency as the original wave is called fundamental wave and the sine waves of higher frequencies are called harmonic waves or simply harmonics.

Figure 3.4 Non-sinusoidal waves and harmonics

In Fig. 3.4 (a) have been shown a trapezoidal and a triangular-type non-sinusoidal wave. In Fig. 3.4 b (i) has been shown a non-sinusoidal wave which is the sum of two component sine waves of different frequencies. One has the same frequency as the non-sinusoidal wave. This is called the fundamental. The other harmonic wave has twice the frequency as the fundamental. This is called the second harmonic. The non-sinusoidal wave shown in 3.4 b (ii) and (iii) are composed of a fundamental wave and a third harmonic. A third harmonic wave has three times the frequency as the fundamental wave. The number of harmonics present in an alternating non-sinusoidal quantity will depend upon the complexity of the wave shape. A symmetrical wave is the one whose positive half is identical to its negative half. Whether a wave is symmetrical or not can be tested by lifting the negative half and shifting it to the positive half axis and placing it just over the positive half. If both the half waves match each other, the wave shape is symmetrical. When generators are built physically symmetrical, the EMF wave shape induced in the coils in such machines are symmetrical in nature. A symmetrical wave will contain fundamental and odd harmonics only. The presence of even harmonics, i.e., 2nd harmonic, 4th harmonic, etc. will be there in non-symmetrical, non-sinusoidal ac waves.

#### 3.1.5 Concept of Average Value and Root Mean Square (RMS) Value of an Alternating Quantity

For a symmetrical alternating voltage or current wave, the positive half is identical to the negative half, and hence the average value of the quantity for a complete cycle is zero. In earlier days the usefulness of such ac was questioned and only dc was considered effective. However, it was observed that when ac is allowed to pass through a resistance element, heat is produced. The question that arose was that if the average value of an alternating quantity is zero, why then was it producing heat. The concept of effective value was then brought in from the point of view of heat equivalence.

#### Average value

The average value of an alternating quantity is the sum of all its values divided by the total number of values. A waveform has continuous variable values with repeat to time, t or angle θ where θ = ωt. The pattern of wave repeats after every cycle. The sum of all the values in a cycle is determined by the integration of its values over a period of time. A full cycle is formed in 2π radians or in T seconds where T is the time period. A symmetrical wave is one where the positive half cycle is exactly the same as the negative half cycle. If we integrate the values for a complete cycle and take its average over one cycle, the quantity becomes equal to zero. The average value if calculated over a complete cycle would become zero.

or

#### Average value of a sinewave or any other symmetrical wave over a complete cycle is zero

For half-wave or full-wave rectified waves we need to calculate the average value. When we intend to calculate the average value of such waves, we calculate the average value for one-half cycle.

The average value is calculated as

or

#### Effective or RMS value

The effective value or RMS value of an alternating quantity is determined by considering equivalent heating effect.

The effective value of an alternating quantity (say current) is that the value of dc current which when flowing through a given circuit element (say a resistance element) for a given time will produce the same amount of heat as produced by the alternating current when flowing through the same circuit element for the same time.

Let I be the equivalent effective value of the ac flowing through a resistance element R for a time t, then the amount of heat produced, H is expressed as

Now, let the alternating current i, be passed through the same resistance R for the same time t, as shown in Fig. 3.5. Current i has been shown divided into n intervals and the magnitudes are i1, i2, i3, etc. Heat produced in t seconds by the ac is equal to the sum of heat produced in n intervals of time during the time t. This can be expressed as

Figure 3.5 RMS value of an alternating current illustrated

Equating expressions (3.3) and (3.4),

or

Thus, the effective value is equal to the square root of the mean of the squares of instantaneous values of the alternating quantities. Alternately, this can be read as square mean root value or root mean square value, i.e., RMS value. While expressing alternating quantities we always use RMS values and write in capital letters as E, I, V, etc. To further clarify the concept of the effective or RMS value let us find out the value of dc current, I which gives the same amount of heating as that of ac when it flows through a resistance of value, say, R. The alternating quantity is represented as, say, i = Im Sin ωt.

The power dissipated in R by the dc current,

The instantaneous value of power dissipated in R by the ac current,

The average value of the second term is zero as it is a cosine function varying with time.

The average value of power, P for the ac current,

Equating the power dissipated due to dc current, I and the ac. current, i we can get the effective value as

or

The effective or RMS value of an alternating quantity is either for one-half of a cycle or for a full cycle as

The RMS values of an alternating quantity of any type of wave shapes can be calculated using analytical methods.

#### 3.1.6 Analytical Method of Calculation of RMS Value, Average Value, and Form Factor

Suppose we have a sinusoidal alternating current, we have to first square it, then take its mean over one cycle or half cycle, and then take the square root (note that RMS value is calculated by making reverse operation i.e., first square, then take mean and then take square root).

Square of the current i = Im sin θ is Im2 sin2θ

Its mean over one cycle is calculated by integrating it from 0 to 2π and dividing by the time period of 2π as follows:

Mean of square

RMS value,

To calculate, let us put

then

ie,

If we calculate the RMS value for half cycle, it can be seen that we will get the same value by calculating as

#### Average value

Average value of a sinusoidally varying quantity over one cycle is zero because for the first half cycle current flows in the positive direction and for the second half cycle same current flows in the negative direction, i.e., in the opposite direction.

Average value has to be calculated by considering one-half cycle as

Similarly average value for sinusoidal voltage, Vav = 0.637 Vm

#### Form factor

As the name suggests, form factor is an indicator of the shape or the form of the ac wave. It is the ratio of the RMS value to the average value of an alternating quantity. For a sinusoidal varying quantity, the form factor Kf is

The sharper the wave shape, the more will be the value of the form factor. For example, for a triangular wave, form factor will be more than 1.11 and for a rectangular wave form factor will be less than 1.11 (in fact, its value will be 1). The peak or crest factor is the ratio of peak or maximum value to its rms value.

It is obvious that by knowing the value of the form factor, the RMS value can be calculated if the average or mean value is known.

#### 3.1.7 RMS and Average Values of Half-wave-rectified Alternating Quantity

A half-rectified sine wave is shown in Fig. 3.6. A half-wave-rectified quantity, whether voltage or current will have its one half cycle blocked by using a diode rectifier as shown. Since the diode allows current to flow in one direction only, current through the load resistance will flow, in one direction only. During the negative half cycle of the input voltage the diode will block current flow and hence no voltage will be appearing across the load during all negative half cycles. For half-wave-rectified current or voltage, we have to consider the current or voltage which is available for the positive half cycles and average it for the complete cycle. For a complete cycle, i.e., from 0 to 2π, current flows only from 0 to π. To calculate the RMS value we have to square the current, take its sum from 0 to π and then take the average for the whole cycle, i.e., from 0 to 2π. Thus the RMS value for a half-wave-rectified current, is calculated as

Figure 3.6 Half-rectified sine wave

[since 2 sin2θ =1 – cos 2θ]

Thus, the RMS value of a full sine wave is and for a half wave, .

Average value of current for half sine wave is

Note that for a complete sine wave, the average value was calculated as

and for a half-rectified sine wave, the average value has been calculated as

Obviously, we note that for a half-rectified wave, the average value is half of that of a full sine wave.

Form factor for a half sine wave quantity is

#### 3.1.8 Concept of Phase and Phase Difference

The position of a coil or a set of coils forming a winding with respect to some axis of reference is called its phase. If three coils are placed at different angles with respect to the reference axis, there exists a space phase difference between these three coils AAʹ, BBʹ, and CCʹ,. When EMFs will be induced in these coils due to the cutting of the magnetic flux or due to change in flux linkages, the EMFs will have similar time phase difference between them as shown in Fig. 3.7.

A magnet has been shown rotating in the anticlockwise direction. Maximum flux will be cut by the coil AAʹ at time, t = 0. Hence, maximum voltage will be in the coil AAʹ at time, t = 0 as has been shown as vA in Fig. 3.7 Maximum flux will be cut by the coil BBʹ after an elapse of angle 30°, i.e., by the time the rotating magnet rotates by an angle of 30°. Similarly, maximum flux will be cut by the coil CCʹ after an elapse of time represented by 60°.

Figure 3.7 Concept of phase and phase difference illustrated

The voltage waves in coil AAʹ, BBʹ and CCʹ will, therefore, have a time phase difference of 30°. (30° corresponds to the time taken by the rotating magnet to rotate by 30°). Since voltage vA is appearing earlier than vB, vA is said to be leading voltage vB. Voltage induced in the three coils AAʹ, BBʹ and CCʹ will have a time phase difference of 30°.

Such phase difference may exist between the voltage and current in an electrical circuit. If current in a circuit changes in accordance with the voltage, i.e., when the voltage is at its maximum value, the current is also at its maximum value, and when the voltage starts increasing in the positive direction from its zero value, the current also starts increasing in the positive direction from its zero value; then, the voltage and current are said to be in phase as shown in Fig. 3.8 (a). Note that the magnitudes of voltage and current may be different. In Fig. 3.8 (b) is shown current, i lagging the voltage by 90°, i.e., by an angle π/2. The expressions for voltage and current as shown in Fig. 3.8 (a) can be written as

Figure 3.8 Phase and phase difference between voltage and current

v = Vm sin ωt

i = Im sin(ωt + 0) = Im sin ωt

The voltage and current shown in Fig. 3.8 (b) can be represented as

v = Vm sin ωt

i = Im sin(ωt – π/2)

If current is leading the voltage by π/2 degrees, we will represent the current, i as

i = Im sin(ωt + π/2)

If two voltages vA and vB are represented as in Fig. 3.8 (c), they can be expressed as

vA = Vm1sin ωt

vB = Vm2sin (ωt + π/2)

This is because voltage vB is leading the voltage vA. The maximum value of vB is appearing π/2 degrees before the maximum value of vA appears. However, if vB is taken as the reference voltage we can express vB and vA as

vB = Vm sin ωt

vA = Vm sin (ωt – π/2)

Example 3.1 An alternating voltage of 100 sin 314 t is applied to a half-wave diode rectifier which is in series with a resistance of 20Ω. What is the RMS value of the current drawn from the supply source?

Solution:

Figure 3.9 Circuit diagram of example 3.1

We have,

v = Vm sin ωt = 100 sin 314 t
Vm = 100 V

For a full sine wave, RMS value

For half-rectified wave

Therefore,

Example 3.2 An alternating sinusoidal voltage of v = 150 sin 100 π t is applied to a circuit which offers a resistance of 50 Ω to the current in one direction and prevents the flow of any current in the opposite direction. Calculate the RMS and average values of the current and the form factor. What is the frequency of the supply?

Solution:

v = 150 sin 100 πt

standard form,

v = Vm sin ωt

The maximum value of voltage,

Vm = 150
ω = 100 sin π = 2πf

frequency,

f = 50 Hz

The circuit in the question is a half-wave-rectified one.

For half sine wave, the RMS value

Therefore,

Form factor,

Example 3.3 Calculate the RMS value, average value and form factor of a half-rectified square voltage shown in Fig. 3.10.

Figure 3.10 Circuit diagram of example 3.3

Solution:

For half-rectified wave,

Here

T = 0.2, v = 10V

Therefore,

Form factor

Example 3.4 Calculate the RMS value and average value of the elevated saw-tooth-type current wave shown in Fig. 3.11

Figure 3.11 Saw-tooth wave of example 3.4

Solution:

It can be seen from the wave shape that 0abcT makes one cycle. The same pattern is being repeated for each time period of T. The equation for the line ab is of the form y = mx + c. Here the slope m is bc/ac, i.e., equal to 5/T. The value of c is 5 and y is represented by i and x by t.

Therefore, the equation of the line ab is

The value of average current is calculated as

(In this case by actual observation of the wave shape as shown in Fig. 3.10, the average value can also be determined)

The RMS value is calculated as

Example 3.5 Find the average value, RMS value and form factor of the saw-tooth current wave shown in Fig. 3.12

Figure 3.12 Diagram for example 3.5

Solution:

The equation of the line ab is of the form

y = mx

Here,

y = i, m = 10/T, and x = t

Therefore, we can write

RMS value of i,

Average value of a right-angle triangle is half of its height, i.e., equal to

Iav = 10/2 = 5A

Form factor

The students are to note that the form factor of a saw-tooth wave has been calculated as 1.15 whereas for a sine wave the value was 1.11. Since a saw-tooth wave is stiffer than a sine wave, its form factor is higher than that of a sine wave.

##### 3.2 SINGLE-PHASE AC CIRCUITS

A resistance, an inductance, and a capacitance are the basic elements of an ac circuit. These elements are connected in series and parallel combinations to form an actual circuit. Circuits may include any two or three elements. For example, we may have one resistance and one inductance connected in series across an ac supply source. We may have one resistance connected in series with one inductance and one capacitance in parallel. Accordingly circuits are named as L-R circuits, L-R-C circuits, etc. We will take up few series circuits, few parallel circuits, and some series–parallel circuits and calculate the main current, branch currents, power, power factor, etc. Before this, we will discuss the behaviour of R, L, and C in ac circuits

#### 3.2.1 Behaviour of R, L, and C in AC Circuits

In this section we will study the relationship of applied voltage and current in an ac circuit involving only a resistance, an inductance, and a capacitance. When a resistance is connected across an ac supply we call it a purely resistive circuit. Similarly an inductance coil connected across an ac supply is called a purely inductive circuit and a capacitance connected across an ac supply is called a purely capacitive circuit. We shall study the phase relationship between the applied voltage and current flowing in each case under steady-state condition.

#### AC applied across a pure resistor

When we say a pure resistance we assume that the resistance wire does not have any inductance or capacitance. Fig. 3.13 shows a pure resistance connected across an ac supply. The voltage and current wave forms as well as the phasor diagram showing the positions of voltage and current have been shown. The instantaneous value of

Voltage, v of the source is v = Vm Sin ωt

Where, Vm is the maximum value of the voltage in Volts; ω = 2πf rad/sec; and f is the frequency of supply voltage in cycles per second.

The current flowing though the circuit will be

At

Figure 3.13 (a) Resistive circuit with a sinusoidal voltage source; (b) voltage and current wave shapes; (c) phasor diagram

The maximum value of i is Im

Therefore,

Thus, I can be written as

or,

i = Im sinωt

The steady-state response of the circuit is also sinusoidal of the same frequency of the voltage applied. As shown in Fig. 3.13 (b), both voltage and current wave shapes are sinusoidal and their frequency is also the same. Since current is proportional to the voltage all the time, the two wave forms are in phase with each other.

The phasor diagram is drawn with the RMS values of the time-varying quantities. As shown in Fig. 3.13 (c), V and I are the RMS values of voltage and current. They have been shown in phase. For the sake of clarity only, the two phasors have been shown with a gap between them.

In a purely resistive circuit, current and voltage are in phase. Power is the product of voltage and current. The product, P = VI has been calculated for all instants of time and has been shown in Fig. 3.13 (b). Power in a resistive circuit is taken as the average power which is

Power factor is the cosine of the phase angle between voltage and current.

In a resistive circuit the phase difference between voltage and current is zero, i.e., they are in phase. So the phase angle θ = 0.

Power factor,

P f = cos θ = cos 0° = 1

#### AC applied across a pure inductor

A pure inductor means that the resistance of the inductor coil is assumed to be zero. The coil has only inductance, L. Such an inductor is connected across a sinusoidally varying voltage, v = Vm sin ωt as has been shown in Fig. 3.14 (a).

As a result of application of voltage, v an alternating current, i will flow through the circuit. This alternating current will produce an alternating magnetic field around the inductor. This alternating or changing field flux will produce an EMF in the coil:

Figure 3.14 (a) Inductive circuit with a sinusoidal voltage input; (b) wave shapes of voltage, current, and power; (c) phasor diagram

This EMF will oppose the voltage applied (remember Lenz’s law). Therefore, we can write

or,

L di = v dt = Vm sin ωt dt

or,

Integrating

or,

or,

where

Thus, we observe that in a purely inductive circuit

v = Vm sin ωt

and

= Im sin (ωt - 90°)

The current, i is also sinusoidal but lagging behind, v by 90̊. The voltage and current wave shapes have been shown in Fig. 3.14 (b). The instantaneous power, p is the product of v and i. The wave shape of instantaneous power has also been shown in the figure. The phasor diagram of RMS values of v and i has been shown in Fig. 3.14 (c). In a purely inductive circuit current, I lags the voltage, V by degrees, i.e., 90°.

Power factor,

cosϕ = cos 90° = 0

Average power

= 0

Hence, the average power absorbed by a pure inductor is zero.

The opposition to current is ωL. This is called inductive reactance, XL which is XL = ωL = 2πfL Ω.

The opposition offered by an inductor to the flow of current is XL which is equal to ωL This is called the inductive reactance and is expressed in Ohms. Inductance, L is expressed in Henry.

As mentioned earlier, the values of alternating quantities are expressed in terms of their effective or RMS values rather than their maximum values.

Therefore,

or,

or,

V = I XL

If V is taken as the reference axis, we can represent V as a phasor and represent as V ∠0°.

Since current, I is lagging voltage, V by 90°, we represent the current as I ∠ –90° or -jI for a purely inductive circuit. Again, if I is taken as the reference axis, then I and V can be represented as I ∠0°. and V ∠ +90° or +jV, respectively, as shown in Fig. 3.15.

Figure 3.15 Phasor diagram of V and I in a purely inductive circuit

Note that j is an operator which indicates rotation of a phasor by 90° in the anti clockwise direction from the reference axis.

Now let us examine why the power absorbed by a pure inductive circuit is zero. We refer back to Fig. 3.14 (b) where it is seen that for one half cycle power is negative and for the next half cycle power is positive. The average value for a complete cycle, the power consumed is zero. Positive power indicates that power is drawn by the circuit from the supply source. When current rises in the circuit, energy is required to establish a magnetic field around the inductor coil. This energy is supplied by the source and is stored in the magnetic field. As the current starts reducing, the magnetic field collapses and the energy is returned to the supply source. Thus, in one half cycle power is drawn by the inductor and in the next half cycle power is returned to the source. This way the net power absorbed by the inductor becomes zero. The power which is being circulated from the source to the inductor and back to the source is called reactive power which will also be discussed in a separate section.

#### AC applied across a pure capacitor

A sinusoidal voltage source has been shown connected across a pure capacitor in Fig. 3.16 (a). When current starts flowing, the capacitor starts getting charged. The charge, q of the capacitor in terms of capacitance of the capacitor, C and supply voltage, v is expressed as

q = Cv

Current, i is the rate of flow of charge. Therefore,

= ωC Vm cos ωt

or,

or,

where,

Hence, in a pure capacitive circuit, v = Vm sin ωt and current Current leads the voltage by 90°.

is called the capacitive reactance of the capacitor.

To express in terms of RMS values,

We take

Figure 3.16 (a) Pure capacitive circuit; (b) wave shapes of voltage, current, and power; (c) phasor diagram

Therefore,

can be written as

Xc is the opposition offered by the capacitor to the flow of current and is called capacitive reactance.

Like an inductor, in a capacitor also the average power absorbed for a complete cycle is zero. When voltage is applied, the capacitor starts getting charged, energy gets stored in the capacitor in the form of electro-static field. When the applied voltage starts falling from its maximum value, the energy starts getting returned to the supply. This way, the power is absorbed from and then returned to the supply source The net power absorbed by a pure capacitor is zero. Since current leads the voltage by 90°, the power factor of the circuit is

P.f = cosϕ = cos 90° = 0

The average or net power in a pure capacitor circuit can be calculated as

or,

= 0

Hence, it is proved that the average power obsorbed by a pure capacitor is zero.

Example 3.6 An inductor of 0.5 H is connected across a 230 V, 50 Hz supply. Write the equations for instantaneous values of voltage and current.

Solution:

V = 230 V, Vm = V = 1.414 × 230 = 324V
XL = ωL = 2π fL = 2 × 3.14 × 50 × 0.5 Ω = 157 Ω
Im = I = 1.414×1.46=2.06 A

The equations are,

V = Vm sin ωt = 324 sin ωt = 324 sin 2π ft = 324 sin 314t

and

Example 3.7 A 230 V, 50 Hz sinusoidal supply is connected across a (i) resistance of 25 Ω ; (ii) inductance of 0.5 H; (iii) capacitance of 100 μF. Write the expressions for instantaneous current in each case.

Solution:

given

V = 230 V
Vm = V = 1.414×230 = 324.3V
ω = 2πf = 2×3.14×50 = 314 rad/sec

Voltage equation is

v = Vm sin ωt

or

v = 324.3 sin 314t

Inductive reactance,

XL = ωL = 314×0.5 = 157 Ω

Capacitive reactance,

When the voltage is applied across a 25 Ω resistor, the current will be

or,

i = 12.97 sin 314t A

Current through the inductor is

or,

i = 2.06 sin (314t - 90°)A

Current through the capacitor is

or,

i = 10.07 sin (314t + 90°)A

Exapmle 3.8 An alternating voltage of RMS value 100 V, 50 Hz is applied separately across a resistance of 10 Ω, an inductor of 100 mH, and a capacitor of 100 μF. Calculate the current flow in each case. Also draw and explain the phasor diagrams.

Solution:

R = 10 Ω
XL = ωL = 2πfL = 2×3.14×50×100×10-3
= 31.4 Ω

Current through R,

Current through I,

Current through C,

We know that in a resistive circuit current is in phase with the applied voltage; in a purely inductive circuit current lags the voltage by 90°; and in a purely capacitive circuit current leads the voltage by 90°. The phasor diagrams have been shown in Fig. 3.17.

Figure 3.17 Phasor diagrams (a) resistive circuit; (b) purely inductive circuit; (c) purely capacitive circuit

#### 3.2.2 L –R Series Circuit

Let us consider a resistance element and an inductor connected in series as shown in Fig. 3.18. A voltage, V of frequency, f is applied across the whole circuit. The voltage drop across the resistance is VR and across the inductor is VL. Current flowing through the circuit is I.

Figure 3.18 (a) R–L series circuit; (b) phasor diagram

VR = IR, VL = IXL where XL = ωL = 2πfL

We have to add VR and VL to get V. But these are to be added vectorially as they are all not in phase, i.e., these vectors are not along the same direction. To draw the current and voltage phasor we take the current I as the reference phasor as shown in Fig. 3.18 (b), since current I is common to VR and VL, i.e., since the same current is flowing through both resistance and inductance. We have, therefore, chosen I as the reference phasor. Voltage drop across the resistance and the current flowing through it are in phase. This is because, as we have seen earlier that in a resistive circuit, voltage and current are in phase. The current flowing through an inductor lags the voltage across it by 90°. That is to say, voltage drop across L, i.e.,VL will lead the current by 90°. Again VL = IXL and XL = ωL. The vector sum of VR and VL is equal to V. The angle between V and I is called the power factor angle ϕ. Power factor is cosϕ. Considering the triangle ABC we can express

or,

or,

where

Z is called the impedance of the total circuit. Triangle ABC in Fig. 3.18 (b) is also called the impendance triangle which is redrawn as in Fig. 3.19. From the impedance triangle

Figure 3.19 Impedance triangle for R–L circuit

or

Z = R + jXL

Where j indicates rotation by 90° in anti-clockwise direction.

or

Z cosϕ = R

and

Z sinϕ = XL

Fig. 3.19 (a) is the same as Fig. 3.19 (b). The current, I has been kept aside which is common to all the sides. Impedance Z can be represented as the vector sum of R and XL. since IXL is leading I by 90 and R is in phase with I, we can write

Power is,

P = VI cosϕ

or,

Power = Volt–Ampere Power × factor

#### Apparent power (S)

It is defined as the product of the RMS value of voltage (V) and current (I). It is denoted by S

Apparent power,

S = VI = Voltage × Current

#### Real or true power or active power (P or W)

It is the power which is actually dissipated in the circuit resistance. (watt-full power)

Or,

P = VI cosϕ Watts or kW

#### Reactive power (Q)

It is the power developed in the inductive reactance of the circuit. (watt-less power)

Q = I2 XL = I2 Z sinϕ = I (ZI) sinϕ

These three powers are shown in the power triangle of Fig. 3.21 (b) from where it can be seen that

S2 = P2 + Q2

#### 3.2.4 Power in an AC Circuit

Let us now develop a general expression for power in an ac circuit by considering the instantaneous values of voltage and current. A sinusoidal voltage is expressed as

Figure 3.20 Wave forms of voltage, current, and power in an R–L series circuit

v = Vm sin ωt

In a circuit when current is lagging the voltage by an angle ϕ, current i is expressed as

i = Im sin (ωt – ϕ)

Sinusoidal waveforms of voltage and current are shown in Fig. 3.20. It is seen that the current wave is lagging the voltage wave by an angle, ϕ which is the power factor angle.

Fig. 3.20 clearly shows that current in an R–L circuit lags voltage by an angle ϕ, which is called the power factor angle.

The expression for the voltage and current in series R–L circuit is,

v = Vm sin ωt
i = Im sin (ωt – ϕ) as I lags V

The power is product of instantaneous values of voltage and current,

p = v × i
= Vm sin ωt×Im sin(ωt – ϕ)

The average power over a complete cycle is calculated as

Now, the second term is a cosine term whose average value over a complete cycle is zero.

Hence, the average power consumed is

Figure 3.21 Power triangle diagram

Pav = Vrms × Irms cosϕ = VI cosϕ W

#### Power factor

It may be defined as the cosine of the phase angle between the voltage and current; cosϕ is known as power factor. Power factor can also be expressed as the ratio, R/Z = resistance/impedance = cosϕ

In Fig. 3.21, the power triangle diagram has been developed from the simple voltage–current relationship in an R–L series circuit. First we have shown I laggingV by the power factor angle ϕ. The inphase component of I is I cosϕ and quadratuse component is I sinϕ as have been shown in Fig. 3.21 (a).

Multiplying all the sides of the triangle ABC by KV (kilo-volt), we can draw the power triangle as in Fig. 3.21 (b)

kVA cosϕ = kW
kVA sinϕ = kVAR

In the power triangle diagram, if ϕ is taken as zero, i.e., if the circuit is resistive, reactive power, Q becomes zero. If the circuit is having pure inductance or capacitance, ϕ = 90, active power, P becomes zero. Reactive power will be present whenever there is inductance or capacitance in the circuit. Inductors and capacitors are energy-storing and energy-releasing devices in the form of magnetic and electric fields, respectively, and are of importance in the field of electrical engineering.

#### 3.2.5 R–C Series Circuit

Consider a circuit consisting of a pure resistance R and connected in series with a pure capacitor C across an ac supply of frequency f as shown in Fig. 3.22.

When the circuit draws a current I, then there are two voltage drops.

1. Drop across pure resistance VR = I × R
2. Drop across pure capacitance VC = I × XC

where and I, VR, VC are the RMS values.

The phasor diagram for such a circuit can be drawn by taking the current as a reference phasor represented by OA as shown in Fig. 3.23. The voltage drop VR across the resistance is in phase with current and is represented by OB. The voltage drop across the capacitance VC lags the current by 90° and is represented by BC. The phasor OC is the phasor sum of two voltages VR and VC. Hence, the OC represents the applied voltages. Thus, in a capacitive circuit, current leads the voltages by an angle ϕ. The same phasor diagram can be drawn by taking voltage, V as the reference vector as shown in Fig. 3.23 (b).

Figure 3.22 R–C series circuit

In Fig. 3.23 (b), we have drawn V as the reference vector. Then current, I has been shown leading V by an angle ϕ. The voltage drop across the resistance, VR = IR has been drawn in phase with I. The voltage drop across the capacitance VC = IXC has been drawn lagging I by 90° (VC lagging I is the same as I leading VC).The length of VR and VC are such that they make an angle of 90°.

In an R–C series circuit, I leads V by an angle ϕ or supply voltage V lags current I by an angle ϕ as shown in the phasor diagram in Fig. 3.23 (b).

Applied voltage,

V = lZ

where

impedance of thecircuit

Voltage and current wave shapes of this circuit are shown in Fig. 3.24, which shows that the current in a capacitive circuit leads the voltage by an angle ϕ, which is called the power factor angle.

Figure 3.23 Phasor diagrams of R–C series circuit

Figure 3.24 Wave forms of voltage and current and their phase relationship in an R–C series circuit

#### Power and power triangle

The expression for voltage and current is

v = Vm Sin ωt
i = Im Sin (ωt + ϕ) as I leads V

Power is the product of voltage and current. The instantaneous power is

P = v × i
= Vm sin ωt × Im sin (ωt + ϕ)
as cos(–ϕ) = cos ϕ

The second term is a cosine term whose average value over a complete cycle is zero. Hence, average power consumed by the circuit is

Pav = Vrms Irms cos ϕ = VIcos ϕ W

The power triangle has been shown in Fig. 3.24 (b).

Thus, various powers are

Apparent power S = VI Volt Amperes or VA

Active power P = VI cosϕ W

Reactive power Q = VI sinϕ VAR

where cosϕ =Power factor of the circuit.

Note: Power factor, cosϕ is lagging for an inductive circuit and is leading for a capacitive circuit.

#### 3.2.6 R–L–C Series Circuit

Consider a circuit consisting of resistance R, inductance L, and capacitance C connected in series with each other across an ac supply. The circuit has been shown in Fig. 3.25.

Figure 3.25 (a) R–L–C series circuit; (b) phasor diagram

The circuit draws a current I. Due to flow of current I, there are voltage drops across R, L, and C which are given by

1. drop across resistance R is VR = IR
2. drop across inductance L is VL = IXL
3. drop across capacitance C is VC = IXC

where I, VR, VL, and VC are the RMS values.

The phasor diagram depends on the magnitude of VL and VC, which obviously depends upon XL and XC. Let us consider the different cases.

(a) When XL > XC, i.e., when inductive reactance is more than the capacitive reactance.

The circuit will effectively be inductive in nature. When XL > XC, obviously, IXL, i.e., VL is greater than IXC, i.e., VC. So the resultant of VL and VC will be VL – VC so that V is the phasor sum of VR and (VL – VC). The phasor sum of VR and (VL - VC) gives the resultant supply voltage V. This is shown in Fig. 3.25 (b) and again redrawn as in Fig. 3.26.

Applied voltage is

or,

V= IZ

Figure 3.26 Phasor diagram of current and voltage drops in an R–L–C circuit where XL > XC

where

Note when XL > XC, the R–L–C series circuit will effectively be an inductive circuit where current I will lag the voltage V as has been shown in the phasor diagram of Fig. 3.26.

(b) When XL < XC

The circuit will effectively be capacitive in nature. When XL < XC, obviously, IXL, i.e., VL is less than IXC, i.e., VC. So the resultant of VL and VC will be directed towards VC. Current I will lead (VC – VL). The phasor sum of VR and (VC – VL) gives the resultant supply voltage V. This is shown in Fig. 3.27.

Figure 3.27 Phasor diagram of an R–L–C series circuit when XL < XC

Applied voltage represented by OB

or,

V = IZ

where

Phase angle,

(c) When XL = XC

When XL = XC, obviously, VL = VC. So, VL and VC will cancel each other and their resultant will be zero. So, VR = V. In such a case the overall circuit will behave like a purely resistive circuit. The phasor diagram is shown in Fig. 3.28. The impedance of the circuit will be minimum, i.e., equal to R.

Figure 3.28 Phasor diagram of an R–L–C series circuit when XL = XC

#### Power and power triangle

The average power consumed by the circuit is

Pav= Average power consumed by R + Average power consumed by L + Average power consumed by C.

But pure L and C never consume any power.

Therefore, Pav = Power taken by R = I2R = IVR

But VR = V cosϕ in all the cases.

Therefore, P = VI cosϕ W.

Thus, for any condition, that is when XL > XC or XL < XC or XL = XC, power can be expressed as P = Voltage × Component of I in phase with V = VI cosϕ

Note that when XL = XC, the component of I in phase with V is I only because I cosϕ = I (as cosϕ = 1).

#### 3.2.7 AC Parallel Circuits

Parallel circuits are formed by two or more series circuits connected to a common source of supply. The parallel brances may include a single element or a combination of elements in series.

Methods for solving ac parallel circuits:

The following three methods are available for solving ac parallel circuits:

1. phasor or vector method
3. using vector algebra (symbolic method or j-operator method)

These methods are explained with examples as follows.

#### 1. Phasor or vector method

A parallel circuit consisting of three branches has been shown in Fig. 3.29. Branch 1 consists of R1, L1, and C1 in series. Branch 2 is resistive and capacitive and branch 3 is resistive and inductive. Let the current be I1, I2, and I3 in the branch 1, 2, and 3, respectively. The total current drawn by the circuit is the phasor sum of I1, I2, and I3.

#### Branch 1

Impedance of branch 1,

Current

I1 = V/Z1

Phase difference of this current with respect to the applied voltage is given by

This current will lag the applied voltage by an angle ϕ1, if XL1 > XC1.

In case XC1 > XL1, then I1, will lead V.

Figure 3.29 AC parallel circuit

#### Branch 2

Impedance of branch 2,

Current                                        I2 = V/Z2

The branch current I2 leads the applied voltage V, by an angle ϕ2, given by

#### Branch 3

Inductive branch 3,

Current

I3 = V/Z3

This current will lag the applied voltage by an angle ϕ3,

Choose a current scale and draw to the scale the current vectors with the voltage as the reference axis. Add vectorially any two currents, say I1 and I2. The vector sum of I1 and I2 is OE as shown in Fig. 3.30 (a). Add vectorially OE with the other branch current, i.e., with I3 to get the sum of the three currents as OF. Convert this length OF to amperes using the current scale choosen earlier.

An alternate method is to show the three currents with the voltage as the horizontal reference axis as shown in Fig. 3.30 (b). Calculate the sum of the horizonal components and vertical components of the currents and then determine the resultant.

The branch currents with their phase angles with respect to V has been shown (not to the scale) separately in Fig. 3.30 (b).

The resultant current I can be found out by resolving the branch currents I1, I2, and I3 into their X and Y components as shown in Fig. 3.30 (b).

X component of I1 (OL) = I1 cos ϕ1

X component of I2 (OM) = I2 cos ϕ2

X component of I3 (ON) = I3 cos ϕ3

Figure 3.30 Phasor diagrams of parallel circuit shown in Fig. 3.29

Sum of X component (active component) of branch currents = I1 cos ϕ1 + I2 cos ϕ2 + I3 cos ϕ3

Y component of I1 (AL) = –I1 sin ϕ1

Y component of I2 (BM) = +I2 sin ϕ2

Y component of I3 (ON) = –I3 sin ϕ3

Sum of Y component (reactive component) of branch currents = –I1 sin ϕ1 + I2 sin ϕ2 – I3 sin ϕ3

Active component of resultant current I = I cosϕ

Reactive component of resultant current I = I sinϕ

Active and reactive components of resultant current must be equal to the sum of active and reactive components of branch currents.

I cosϕ = I1 cos ϕ1 + I2 cos ϕ2 + I3 cos ϕ3
I sinϕ = -I1 sin ϕ1 + I2 sin ϕ2 - I3 sin ϕ3

Resultant current

Resultant current lags the applied voltage if ϕ is –ve, and leads the voltage in case ϕ is +ve.

Power factor of the circuit as a whole is

Concept of Admittance Method: Admittance is defined as the reciprocal of the impedance. It is denoted by Y and is measured in unit mho or siemens.

If the circuit contains R and L,

Z = R + j XL;

If the circuit contains R and C,

Z = R − jXC.

Considering XL and XC as X we can write Z = R ± jX.

Consider an impedance as given by

Z = R ± jX

Positive sign is for an inductive circuit and negative sign is for a capacitive circuit.

Rationalizing the above expression,

Figure 3.31 Impedance and admittance triangles

Y = G ∓ jB

Where G = Conductance

and B = Susceptance

B is negative if the circuit is inductive and B is positive if the circuit is capacitive. The impedance triangle and admittance triangle for the circuit have been shown in Fig. 3.31.

Consider a parallel circuit consisting of two branches 1 and 2. Branch 1 has R1 and L1 series while Branch 2 has R2 and C1 series, respectively. The voltage applied to the circuit is V Volts as shown in Fig. 3.32.

Total conductance is found by adding the conductances of two branches. Similarly, the total susceptance is found by algebraically adding the individual susceptance of different branches.

Total conductance

G = G1 + G2

Total susceptance

B = (–B1) + B2

Total current

I = VY

Power factor

cosϕ = G/Y

Figure 3.32 Parallel circuit

It is quite clear that this method requires calculations which are time consuming. To illustrate this method we will take one example.

Example 3.9 Two impedences Z1 and Z2 are connected in parallel across a 230 V, 50 Hz supply. The impelance, Z1 consists of a resistance of 14 Ω and an inductance of 16 mH. The impedance, Z2 consists of a resistance of 18 Ω and an inductance of 32 mH. Calculate the branch currents, line current, and total power factor. Draw the phasor diagram showing the voltage and currents.

Solution:

Figure 3.33

Let

R1 = 14 Ω, XL = ωLI = 2πfL1 = 2 × 3.14 × 50 × 16 × 10-3 = 5 Ω
R2 = 18 Ω, XL2 = ωL2 = 2πfL2 = 2 × 3.14 × 50 × 32 × 10−3 = 10 Ω

The phase angles of Z1 and Z2 are calculated from the impedance triangles as

Figure 3.34 (a)

Thus,

Admittance of branch I is Y1 and admittance of branch II is Y2

Taking voltage, V as the reference axis,

The phasor diagram showing V, I1, I2 has been shown in Fig. 3.34 (b). The sum of I1 and I2 gives total current, I. The cos of angle between V and I gives the value of total power factor

Figure 3.34 (b)

Taking the cosine and sine components of the branch currents and the line current

I cosϕ = I1 cosϕ1 + I2 cosϕ2

and

I sinϕ = I1 sinϕ + I2 sinϕ2

Substituting values

I cosϕ = 15.41 × 0.942 + 11.15 × 0.335 = 18.24
I sinϕ = 15.41 × 0.325 + 11.15 × 0.485 = 10.4

= 0.57 tan-1 0.57 = 30° Power factor = cosϕ = cos 30° = 0.866 lagging

Current, I can also be calculated as

I = VY

Where

Y = Y1 + Y2

Example 3.10 In Fig. 3.35 is shown a parallel circuit, an inductance L and a parallel R connected across 200 V, 50 Hz ac supply. Calculate:

(a) the current drawn from the supply; (b) apparent power; (c) real power; (d) reactive power.

Figure 3.35

Solution:

Resistance of resistive branch, R = 40 Ω

Inductive reactance of inductive branch.

XL = 2πfL
= 2π × 50 × 0.0637 = 20 Ω

Current drawn by resistive branch,

Current drawn by inductive branch,

Figure 3.36

1. Current drawn from the supply (see Fig. 3.36),
2. Apparent power, S = V × I = 200 × 11.18 = 2.236 kVA
3. Real power, P = V I Cosϕ = V IR = 200 × 5 = 1.0 kW
4. Reactive power, Q = VI Sinϕ = V × IL = 200 × 10 = 2.0 kVAR

Example 3.11 The parallel circuit shown in the Fig. 3.37 is connected across a single-phase 100 V, 50 Hz ac supply. Calculate:

1. the branch currents
2. the total current
3. the supply power factor
4. the active and reactive power supplied by the source.

Figure 3.37

Solution:

It is assumed that the students are aware of the method of representation of a complex number in the forms of a + ib or a + jb. However, this has been explained in the next section.

I = I1 + I2 = 10∠ – 40° + 10∠48°
I = 10 cos 40° – j10 sin 40° + 10 cos 48° + j10 sin 48°
= (10 cos 40° + 10 cos 48°) + j (10 sin 48° – 10 sin 40°)
= 10 × 0.766 + 10 × 0.669 + j(10 × 0.743 – 10 × 0.642)
= 7.66 + 6.69 + j(7.43 – 6.48)
= 14.35 + j0.95
= 14.45A

Power factor angle,

Power factor,

cos ϕ = 0.99

Active power

= VI cos ϕ
= 100 × 14.45 × 0.99
= 1330 W

Reactive power

= VI sin ϕ
= 100 × 14.45 × 0.069
= 99.7 VAR

#### 3. Use of phasor algebra

Alternating quantities like voltage, current, etc., can be represented either in the polar form or in the rectangular form on real and imaginary axis. In Fig. 3.38 is shown a voltage, V represented in the complex plane.

The voltage, V can be represented as V This is called the polar form of representation. Voltage V can also be represented as V = a + jb = V cosϕ + jV sinϕ. This is called the rectangular form of representation using a j operator.

#### Significance of operator j

The operator j used in the above expression indicates a real operation. This operation when applied to a phasor, indicates the rotation of that phasor in the counter-clockwise direction through 90° without changing its magnitude. As such it has been referred to as an operator. For example, let a phasor A drawn from O to A be in phase with the X-axis as has been shown in Fig. 3.39 (a). This phasor when represented by jA shows that the phasor A has been rotated in the anticlockwise direction by an angle 90° and as such its position now is along the Y-axis. If the operator j is again applied to phasor jA, it turns in the counter-clockwise direction through another 90°, thus giving a phasor j2A which is equal and opposite to the phasor A, i.e., equal to –A. See Fig. 3.39 (a).

Figure 3.38 Representation of a phasor

Thus, j2 can be seen as equal to –1. Therefore, the value of j becomes equal to .

Hence,

j = + 90° CCW rotation from OX-axis
j2 = j × j = ()2 = – 1, 180° CCW rotation from OX-axis
j3 = ()3 = – , 270° CCW rotation from OX-axis

and

j4 = ()4 = (–1)2 = 1, 360° CCW rotation from OX-axis

From above, it is concluded that j is an operator rather than a real number. However, it represents a phasor along the Y-axis, whereas the real number is represented along the X-axis.

As shown in Fig. 3.39 (b), phasor OB can be represented as 5 in the polar form. In the rectangular form OB is represented as 4 + j3

Figure 3.39 Use of operator j to represent a phasor

Figure 3.40 Addition of phasor quantities

= 5 × 0.8 + j 5 × 0.6 = 4 + j3

#### Addition and subtraction of phasor quantities

Refer to Fig. 3.40.

Let

V1 = a1 + jb1 and V2 = a2 + jb2

V = V1 + V2
= (a1 + jb1) + (a2 + jb2)
= (a1 + a2) + j(b1 + b2)

Magnitude of the resultant vector

Phase angle,

Subtraction:

V = V1 – V2
= (a1 + jb1) – (a2 – jb2)
= (a1 – a2) + j(b1 – b2)

Magnitude of the resultant vector

Phase angle,

#### Multiplication and division of phasor quantities

Let

V1 = a1 + jb1 = V1∠θ1; where
V2 = a2 + jb2 = V2 ∠θ2; where

#### Multiplication

V1 = V1 × V2 = V1∠θ1 × V2∠θ2 in the polar form
= V1 V2 ∠(θ1 + ∠θ2), angles are added algebraically

#### Division

angles are subtracted algebraically

Example 3.12 A coil having a resistance of 5 Ω and inductance of 30 mH in series are connected across a 230 V, 50 Hz supply. Calculate current, power factor, and power consumed.

Solution:

R = 5 Ω
L = 30 × 10–3 H

Inductive reactance,

XL = ωL = 2πfL
XL = 2 × 3.14 × 50 × 30 × 10–3
= 9.42 Ω

Impedance,

Z = R + jXL = 5 + j9.42
= 10.66 ∠tan-1 1.884
= 10.66 ∠62° Ω

Current,

Magnitude of

I = 21.57 A

Current I is lagging the voltage, V by 62°. Power factor =cosϕ

Here, p.f. = cos 62° = 0.47 lagging. The phasor diagram along with its circuit has been shown in Fig. 3.41.

Power consumed

= VI cosϕ
= 230 × 21.57 × 0.47
= 2331.7 W

Figure 3.41

Example 3.13 For the R–L–C series circuit shown in Fig. 3.42 (a), calculate current, power factor, and power consumed.

Figure 3.42 (a)

Solution:

Inductive reactance,

XL = 2πfL = 2 × 3.14 × 50 × 0.15
= 47.1 Ω

Capacitive reactance,

= 31.84 Ω

Here, XL is greater than XC. Therefore, the circuit reactance will be XL–XC.

Impedance,

Z = R + J(XL–XC)
= 15 + j(47.1 – 31.84)
= 15 + j15.26 Ω

Current,

= 10.75∠-45.3° A

This shows that magnitude of current is 10.75 A and the current lags the voltage by 45.3°.

Power factor,

cosϕ = cos 45.3° = 0.703 lagging

Power consumed,

P = VI cosϕ
= 230 × 10.75 × 0.703 W
= 1738.16 W

Example 3.14 Two coils having impedance Z1 and Z2 are connected in series across a 230 V, 50 Hz power supply as shown in Fig. 3.42 (b).

Figure 3.42 (b)

The voltage drop across Z1 is equal to 120∠30° V. Calculate the value of Z2.

Solution:

We have,

V = V1 + V2

or

V2 = V – V1 = 230 ∠0 – 120 ∠30°

= 230(cos 0° + j sin 0°) – 120(cos 30° + j sin 30°)

= 230 – 120 × 0.866 – j120 × 0.5
= 139.6∠–25.4°

Since this is a series circuit, the current flowing through the circuit is the same.

The circuit current can be calculated by using any of the following relations

current,

since the same current will flow through Z2,

= 17.45(cos15.4° - jsin15.4°)
= 17.45 × 0.964 - j17.45 × 0.2656
= 16.82 – j4.6 Ω

Note that the impedance of an R–L circuit is R + j XL and impedance of an R–C circuit is R – jXC. Since, Z = 16.82 – j4.63, it must be an R–C circuit.

Thus, impedance coil Z2 is written as

Z2 = R - jXc

where,

R = 16.82 Ω

and

XC = 4.63 Ω

Substituting

or,

= 687.8 μF

Example 3.15 An alternating voltage, V = (160 + j170)V is connected across an L–R series circuit. A current of I = (12 – j5) A flows through the circuit. Calculate impedance, power factor, and power consumed. Draw the phasor diagram.

Solution:

Impedance,

= 12.13∠46.8° + 22.6°
= 12.13∠69.4° Ω
Z = 12.13(cos 69.4° + j sin 69.4°)
= 12.13 × 0.35 + j 12.13 × 0.93
= 4.24 + j 11.28 = R + jXL

The series circuit consists of a resistance of 4.24 Ω and an inductive reactance of 11.28 Ω. The phasor diagram is drawn by considering a reference axis. Let x-axis be the reference axis. The voltage applied has a magnitude of 233 V and is making 46.8° with the reference axis in the positive direction, i.e., the anticlockwise direction. Current flowing is 19.2 A lagging the reference axis by 22.6° as shown in Fig. 3.43. The angle between phasor V and phasor I is 69.4°. This is the power factor angle.

Power factor,                                      cosϕ = cos69.4°

= 0.35 lagging

Figure 3.43

Power consumed,                        P = VIcosϕ

P = 233 × 19.2 × 0.35

= 1565.76 W

If supply frequency is taken as 50 Hz, the value of L can be calculated from XL.

XL = 11.28 Ω
XL = ωL = 2πf L
= 35mH

Example 3.16 A sinusoidal voltage of v = 325 sin 314t when applied across an L–R series circuit causes a current of i = 14.14 sin (314t – 60°) flowing through the circuit. Calculate the value of L and R of the circuit. Also calculate power consumed.

Solution:

given

v = 325sin 314t

comparing

v = Vm sinωt
Vm = 325 V

RMS value,

ω = 314

or,

2πf = 314

given

i = 14.14 sin(314t – 60°)

comparing

i = Im sin(ωt – ϕ)

RMS value,

Power factor angle,

ϕ = 60°

Power factor,

cosϕ = cos 60° = 0.5 lagging

Power,

P = VI cosϕ = 230 × 10 × 0.5
= 1150 W

Impedance

In complex form,

Z = 23(cos60° + jsin 60°)
= 23 × 0.5 + j23 × 0.866
= 11.5 + j22.99
= R + jXL

Thus, resistance of the circuit,

R = 11.5 Ω

Inductive reactance,

XL = 22.99 Ω

or,

ωL = 22.99
= 73.21 mH

Example 3.17 A variable resistance R and an inductance L of value 100 mH in series are connected across at 50 Hz supply. Calculate at what value of R the voltage across the inductor will be half the supply voltage.

Solution:

Figure 3.44

Given,

L = 100 mH
XL = ωL = 2πfL
= 2 × 3.14 × 50 × 100 × 10–3
= 31.4 Ω

We have to find R for which

Equating,

or,

or,

Squaring,

R2 + X2L = (62.8)2
R2 = (62.8)2 – (31.4)2
R2 = 3943.8 – 985.9
R = 54.4 Ω

Example 3.18 A voltage of v = 100 sin (314t + 0) is applied across a resistance and inductance in series. A current of 10 sin (314t – π/6) flows through the circuit. Calculate the value of R and L of the circuit. Also calculate power and power factor.

Solution:

v = 100 sin 314t
= Vm sin ωt
Vm = 100 V

Similarly,

i = 10 sin (ωt – 30°)
= Im sin (ωt – 30°)

Current, I is lagging V by 30°.

Power factor,

cos ϕ = cos 30°
= 0.866 lagging

Impedance,

In rectangular form,

Z = 10(cos 30° + j sin 30°)

= 10 × 0.866 + j 10 × 0.5

= 8.66 + j5.0

= R + jXL

R = 8.66 Ω

and

XL = 5.0 Ω

Again

XL = ωL = 314L
= 15.92 mH

Example 3.19 The expression of applied voltage and current flowing through an ac series L–R circuit are

Calculate for the circuit (i) power factor; (ii) average power; (iii) impedance; (iv) R and L

Solution:

We will compare the voltage, and current, i with the standard form

v = Vm sin ωt and
I = Im sin ωt

From the data provided

Vm = 200, Im = 20, ω = 314

RMS values,

Figure 3.45

We have represented in Fig. 3.45 the voltage and current with respect to a common reference axis. The voltage, V is leading the reference axis by π/3°, i.e., 60°, while current I is leading the reference axis by 30°. The phase angle between V and I is 30°. The current in the circuit lags the voltage by 30°.

Power factor, cos ϕ = cos 30° = 0.866 lagging

Average power,                P = VI cosϕ = 141.4 × 14.14 × 0.866

= 1732 W

The impedance of the circuit,

Again,

In the polar form,

Expressing in the rectangular form,

Figure 3.46

Z = Z cosϕ + j Z sinϕ

= 10 cos 30° + j Z sin 30°

= 10 × 0.866 + j10 × 0.5

= 8.66 + j 5

= R + j XL

R = 8.66 Ω

XL = ωL

Example 3.20 In an L–R–C series circuit the voltage drops across the resistor, inductor, and capacitor are 20 V, 60 V, and 30 V, respectively. Calculate the magnitude of the applied voltage and the power factor of the circuit.

Solution:

In Fig. 3.47 the voltage drops across the circuit components and their phase relationship have been drawn. Since it is a series circuit, it is always convenient to take current as the reference axis. The voltage drop across the resistor and the current are in phase. Voltage across the inductor will lead the current, and voltage across the capacitor will lag the current. The circuit diagram and the phasor diagram have been shown.

Figure 3.47

From triangle ABC

AC2 = AB2 + BC2

or,

V2 = (VR)2 + (VL– VC)2

= (20)2 + (60 – 30)2

= 1300

or,

V = 36 V

Power factor

Power factor angle,

ϕ = 56.6°

Since current I is lagging V by an angle ϕ = 56.6°, the power factor is taken as lagging power factor.

Example 3.21 In the circuit shown in Fig. 3.48, calculate the value of R and C.

Figure 3.48

Solution:

v = 325 sin 314t, ω = 314

Current I is leading V by degrees.

Taking V as the reference axis,

In an L–R–C series circuit, if current I is leading the voltage V, we have to consider the circuit as leading p.f circuit. This means the capacitive reactance is more than the inductive reactance (i.e., the circuit is effectively an R–C circuit. We will draw the phasor diagram by taking current on the reference axis. Here we see that V is lagging I by the power factor angle. That is, I is leading V by an angle ϕ. The phasor diagram taking I as the reference axis has been shown in Fig. 3.49

Figure 3.49

If we take V as the reference axis,

Z = 23 cos 30° – j 23 sin 30°

= 23 × 0.866 – j 23 × 0.5

= 19.9 – j 11.5

Z = R – j (XC – XL)

R = 19.9 Ω

XC – XL = 11.5 Ω

XL = ωL = 314 × 100 × 10–3 = 31.4 Ω

XC – 31.4 = 11.5 Ω

XC = 42.9 Ω

or,

Power factor = cos ϕ = cos 30° = 0.866 leading.

In Fig. 3.50, AB = IR = VR has been drawn in the direction of current I. I(Xc – XL) is effectively a voltage drop which is capacitive in nature. I will lead I(Xc – XL), or we can say that I(Xc – XL) will lag I by 90°. BC has been shown lagging AB by 90°. The sum of AB and BC is AC which the total voltage, V and V = IZ. By taking away I from all the sides of the triangle ABC, the impedance triangle has be drawn.

Figure 3.50

Example 3.22 A resistance of 15 Ω and an inductance of 100 mH are connected in parallel across at 230 V, 50 Hz supply. Calculate the branch currents, line current, and power factor. Also calculate the power consumed in the circuit.

Solution:

The circuit diagram and the phasor diagram have been shown in Fig. 3.51. We note that in a parallel circuit the voltage applied across the branches is the same. The current in the resistive branch is in phase with the voltage while current in the inductive branch lags the voltage by 90°. The phasor sum of the branch currents gives us the total line current. Since in a parallel circuit voltage, V is common to the parallel branches, we generally take V as the reference axis while drawing the phasor diagram. Current through the resistive branch, IR has been drawn in phase with V. Current through the inductive branch, IL is lagging V by 90°. The sum of IR and IL gives I as has been shown in Fig. 3.51.

Power factor,

Figure 3.51

Now using the given values, calculations are made as follows.

Inductive reactance,

XL = ωL = 2πfL

= 2 × 3.14 × 50 × 100 × 10–3

= 31.4 Ω

Power factor = lagging

Power factor angle, ϕ = cos–1 0.9 = 25°

Since the line current I is lagging the voltage V by 25°, the power factor is mentioned as lagging. The students should note that while mentioning power factor, it is essential to indicate whether the same is lagging or leading.

Power consumed

P = VI cos ϕ = 230 × 17 × 0.9 = 3519 W

Example 3.23 For the circuit shown in Fig. 3.52 calculate the total current drawn from the supply. Also calculate the power and power factor of the circuit.

Figure 3.52

Solution:

For branch I, the impedance Z1 is calculated as

Z1 = R1 + jXL1 = 5 + jωL1 = 5 + j 2π × 50 × 150 × 10–3
= 5 + j 31.4
= 31.8∠81° Ω

Similarly for branch II,

Z2 = R2 + j XL2 = 50 + j 2π × 50 × 15 × 10–3

= 50 + j 4.71
= 50.22∠5.4° Ω

Current,

Current,

Total Current,             I = I1 + I2 = 7.23∠81° + 4.58∠ – 5.4°

= 7.23cos 81° - j7.23 sin 81° + 4.58 cos 5.4° - j4.58 sin 5.4°

= 7.23 × 0.156 × j7.23 × 0.987 + 4.58 × 0.995 – j4.58 × 0.09

= 1.127 – j7.136 + 4.557 – j0.414

Line current,            I = 9.44∠–53° A

The phasor diagram representing the branch currents and the line current with respect to the supply voltage has been shown in Fig. 3.53. The line current lags the applied voltage by an angle, ϕ = 53°.

Figure 3.53

Thus,

power factor = cos ϕ = cos 53°
= 0.6 lagging

It may be noted that the branch II is more resistive and less inductive than branch I. That is why current I1 is more lagging than current I2.

Power,                            P = VI cos ϕ

= 230 × 9.44 × cos 53°

= 230 × 9.44 × 0.6

= 1302.7 W

Example 3.24 Two impedances Z1 = 10 + j12 and Z2 = 12 – j10 are connected in parallel across a 230 V, 50 HZ supply. Calculate the current, power factor, and power consumed.

Solution:

The two impedances are of the form, Z1 = R1 + jXL and Z2 = R2 – jXC

Z1 is composed of a resistor and an inductor while Z2 is composed of a resistor and a capacitor.

Figure 3.54

Z1 = 10 + j12
= 15.62∠tan-1 1.2
= 15.62∠50° Ω
Z2 = 12 – j10
= 15.62∠40° Ω

We may calculate, and and then add I1 and I2 to get I. Alternately, we may find the equivalent impedance of the circuit, Z and then find,

So,

or,

Total line current,

Current, I lags voltage, V by 15°

Power factor = cos ϕ = cos 15° = 0.96 lagging

Magnitude of current, I = 20.68 A

Supply voltage, V = 230 V

Power consumed,

P = VI cos ϕ

= 230 × 20.68 × 0.96

= 4566 W

= 4.566 kW

Example 3.25 For the circuit shown in Fig. 3.55 calculate the total current, power, and power factor of the whole circuit. Also calculate the reactive power and apparent power of the circuit. Draw the phasor diagram.

Figure 3.55

Solution:

ω = 2πf = 6.28 × 50 = 314

XL = ωL = 314 × 50 × 10−3 = 15.57 Ω

Z1 = R1 + jXL = 12 + j15.57 = 19.6∠52.2°

Z2 = R2 − jXc = 50 − j63.7 = 80.9∠−52°

I = I1 + I2 = 10.2∠−22.2° + 2.47∠82°

= 10.2 cos22.2° − j10.2 sin22.2° + 2.47 cos82° + j2.47 sin82°

= 10.2 × 0.92 − j10.2 × 0.37 + 2.47 × 0.14 + j2.47 × 0.99

= 9.72 − j1.33

Total Current,   I = 9.8 A

The voltage V is making an angle of +30° with the reference axis as shown in Fig. 3.56. Current I2 is making 82° with the reference axis; current I1 is making −22.2° with the reference axis. The resultant of I1 and I2 is I. Current, I is making an angle of −7.8° with the reference axis. The phase difference between V and I is 37.8° as has been shown in Fig. 3.47.

Figure 3.56

Therefore, power factor, cos ϕ = cos 37.8° = 0.79 lagging

Power,                           P = VI cos ϕ

= 200 × 9.8 × 0.79

= 1548.4 W = 1.5484 kW

Reactive Power                 = VI sin ϕ

= 200 × 9.8 × sin 37.8°

= 200 × 9.8 × 0.61

= 1195.6 VARs = 1.1956 kVARs

Apparent Power                = VI

= 200 × 9.8

= 1960 VA

= 1.96 kVA

To check,

(kVA)2 = (kW)2 + (kVAR)
= 1.96

Example 3.26 Three impedances, Z1, Z2, Z3 are connected in parallel across a 230 V, 50 HZ supply. The values are given Z1 = 12∠30°; Z2 = 8∠−30°; Z3 10∠60°. Calculate the total admittance, equivalent impedance, total current, power factor, and power consumed by the whole circuit.

Solution:

Figure 3.57

Total admittance,                              Y = Y1 + Y2 + Y3

Substituting,

= 0.08∠−30° + 0.125∠30° + 0.1∠−60°

= 0.08 (cos 30° − jsin 30°) + 0.125 (cos 30° + jsin 30°) + 0.1 (cos 60° − jsin 60°)

= (0.227 − j 0.064) mho

= 0.235 ∠−14° mho

Impedance,

Total current,

= 54.05 ∠−14° A

Power factor,

cos ϕ = cos14° = 0.97 lagging

P = VI cos ϕ = 230 × 54.05 × 0.97 = 12058 W = 12.058 kW

Example 3.27 For the circuit shown in Fig. 3.58 calculate the current in each branch and total current by the admittance method. Also calculate power and power factor of the total circuit.

Figure 3.58

Solution:

I1 = VY1 = 230 × 0.0589∠−45° = 13.54∠−45° A

I2 = VY2 = 230 × 0.0559∠−64° = 12.85 ∠−64° A

I = I1 + I2 = 13.54∠−45° + 12.85∠−64°

or,                         I = 15.2 − j21 = 25.9 ∠−54° A

Power factor = cos ϕ= cos 54° = 0.58 lagging

Power                         = VI cos ϕ= 230 × 25.9 × 0.58

= 3455 W = 3.455 kW

#### 3.2.8 AC Series—Parallel Circuits

Consider the series–parallel circuit consisting of three branches A, B, and C as shown. In Fig. 3.59.

Figure 3.59

Impedance of branch A,                      ZA = R1 + jX1

Impedance of branch B,                      ZB = R2 + jX2

Impedance of branch C,                      ZC = R3 + jX3

Total impedance of the circuit Z is

Total current,

Current,

(applying current divider rule)

Current,

By applying the admittance method we can also solve the problem as

Total admittance of the parallel branches B and C

YBC = YB + YC

Impedance,

Total impedance,

Z = ZA + ZBC

Total current,

Example 3.28 Determine the total current drawn from the supply by the series–parallel circuit shown in Fig. 3.60. Also calculate the power factor of the circuit.

Figure 3.60

ω = 2πf = 2 × 3.14 × 50 = 314

Z1 = 10 + j314 × 0.0638 = 10 + j20 = 22.36∠64°

Z2 = 8 − jXc

Z2 = 8 − j8 = 11.3∠−45°

Z3 = 6 + j314 × 0.0319 = 6 + j10 = 11.66 ∠59°

Equivalent impedance,

= 11.68∠−15° + 11.66∠59°
= 11.21 − j3 + 6 + j10
= 17.21 + j7
= 18.58∠22° Ω

Current,

Total current,                         I = 12.37 A

Power factor, cos ϕ = cos 22° = 0.92 lagging

Example 3.29 What should be the value of R for which a current of 25 A will flow through it in the circuit shown in Fig. 3.61.

Figure 3.61

Solution:

Z1 = 5Ω ω = 2πf = 2π × 50 = 314

Z2 = 10 + j314 × 50 × 10−3

= 10 + j15.7

= 18.6∠57.5° Ω

Z3 = R Ω

Total impedance,

or,

Z = 4.28∠11° + R

Again,

Equating the above two expressions for Z,

4.28∠11° + R = 9.2

or,                                     R = 9.2 − 4.28∠11°

= 9.2 − 4.28 (cos11° + jsin11°)

= 9.2 − 4.19 + j 0.8

= 5.01 + j 0.8

Considering the real part, R = 5.01 Ω

Example 3.30 In the series–parallel circuit shown in Fig. 3.62, the parallel branches A and B are in series with branch C. The impedances are ZA = (4 + j3) Ω, ZB = (10 − j7) Ω and ZC = (6 + j5) Ω. If the voltage applied to the circuit is 200 V at 50 Hz, calculate (a) current IA, IB, and IC; (b) the power factor for the whole circuit. Draw also the phasor diagram.

Solution:

Figure 3.62

ZA = (4 + j3) = 5∠36.9° Ω

ZB = (10 − j7) = 12.2∠−35° Ω

ZC = (6 + j5) = 7.8∠39.8° Ω

ZA + ZB = 4 + j3 + 10 − j7

= 14 − j4

= 14.56∠−16°

= 4.19∠17.9°
= 4.19(cos 17.9° + jsin17.9°)
= 4 + j1.3
Z = Zc + ZAB = 6 + j5 + 4 +j1.3
= 10 + j6.3 = 11.8∠32.2°

Let

V = 200∠0°

Using current divider rule,

Phase angle between applied voltage V and line current I is − 32.2°

Hence, power factor of the whole circuit = cos ϕ = cos 32.2°

= 0.846 lagging

Voltage drop across series branch C, VC = ICZC

= 16.3∠−32.2° × 7.8∠39.8°
= 127.53∠−7.6° V

Voltage drop across parallel branches, VA = VB = IC · ZAB

= 16.35∠−32.2° × 4.19 ∠17.9
= 68.5∠−14.3° V

Note that voltage across the parallel branches is also equal to IAZA or IBZB.

The complete phasor diagram is shown in Fig. 3.63.

Figure 3.63 Phasor diagram representing voltages and currents in the circuit of Fig. 3.55

Example 3.31 In the circuit shown in Fig. 3.64, determine the voltage at 50 Hz to be applied across terminals AB in order that a current of 10 A flows in the capacitor.

Figure 3.64

Solution:

Z1 + j2π × 50 × 0.0191 = 5 + j6 = 7.81∠50.2° Ω
Z3 = 8 + j2π × 50 × 0.0318 = (8 + j10) Ω = 12.8∠51.34° Ω

Current in the capacitive branch,

I2 = 10∠0° = 10 + j0 A

Voltage drop across the parallel branch

VAC = I2 Z2

Let                                     = 10∠0° × 10.63∠−48.8°

= 106.3∠−48.8° V

= (70.02 − j79.98)V

Current in inductive branch

= (−2.13 − j13.44)A

Circuit current,                                 I = I1 + I2 = 10 + j0 − 2.13 − j13.44 = 7.87 − j13.44

= 15.57∠−59.65° A

Voltage drop across series branch, VCB = IZ3

= 15.57∠−59.65° × 12.8∠51.34°

= 199.4∠−8.31° V

Voltage applied across terminals AB, VAB = VAC + VCB

= (267.33 − j108.8) V

= 288.62∠22.15° V

Example 3.32 In a series–parallel circuit shown in Fig. 3.65, the parallel branches A, B, and C are in series with branch D. Calculate:

1. the impedance of the overall circuit,
2. current taken by the circuit, and
3. power consumed by each branch and the total power consumed.

Figure 3.65

Solution:

1. Impedance of branch A,

ZA = 2 + j0 = 2 Ω

Impedance of branch B,

ZB = 3 + J4

Impedance of branch C, ZC = (2 − j2) Ω

Equivalent impedance of the parallel circuit,

= 1.136 − j0.118 = 1.142 Ω

Impedance of branch D,                                            ZD = 1 + j1

Total impedance of the overall circuit Z, = Zp + ZD = 1.136 − j0.118 + 1 + j1 = 2.136 + j0.882

= 2.311∠22.46° Ω

2. V = 110 V

Z = 2.311 Ω

3. ID = 47.6 A

RD = 1 Ω

Power consumed by branch              D = ID2RD

= (47.6)2 × 1

= 2265.8 W

Voltage drop across terminals PQ = IZP

Current in branch A,

RA = 2 Ω

Power consumed by branch A,

= IA2RA = (27.18)2 × 2 = 1477.5 W

Current in branch B,

RB = 3 Ω

Power consumed by branch B,

=IB2 RB = (10.87)2 × 3 = 354.5 W

Current in branch C,

RC = 2 Ω

Power consumed by branch C,

= IC2 RC = (19.21)2 × 2 = 738 W

Total power consumed by circuit           = 1477.5 + 354.5 + 738 + 2265.8

= 4835.8 W

##### 3.3 RESONANCE IN AC CIRCUITS

In ac circuits resonance occurs when two independent energy storing devices are capable of interchanging energy from one another. For example, inductance and capacitance are the two energy-storing devices or elements of an ac circuit, which may create a condition of resonance.

Resonance occurs in other systems also, like in a mechanical system, where mass and spring are the two energy-storing elements and they may create a condition of resonance. Mass stores energy when in motion and a spring stores energy when it is elongated or compressed.

An electric circuit generally consists of circuit elements like resistance, inductance, and capacitance. The voltage and frequency are generally constant at the supply terminal. However, in electronic communication circuits, the supply voltage may have variable frequency. When frequency is variable, the inductive and capacitive reactance of the circuit elements will change The current in the circuit will depend upon the values of XL and XC and that of R. A condition may occur at a particular frequency that the impedance offered to the flow of current is maximum or minimum. The circuit elements, namely the inductance element and the capacitance element are often connected in series or in parallel. It will be interesting and also useful to study the effect of varying input frequency on the circuit condition when these elements are connected in series or in parallel.

#### 3.3.1 Resonance in AC Series Circuit

Let us consider a series circuit consisting of a resistor, an inductor, and a capacitor as shown in Fig. 3.66 (a). The supply voltage is constant but its frequency is variable.

Figure 3.66 Resonance in R–L–C series circuit: (a) circuit diagram; (b) variation of R, XL, XC with frequency; (c) variation of impedance

The impedance of the circuit,

The current flowing through the circuit is

As the frequency is changing, both XL and XC will change. Inductive reactance XL will increase as the frequency, f is increasing while the capacitive reactance, XC will decrease with increasing frequency. The value of R is independent of frequency. The variation of R, XL, and XC with variation of frequency, f has been shown in Fig. 3.66 (b). It may be noted that inductive reactance is jXL and capacitive reactance is − jXc, i.e., vectorially they should be shown in opposite directions. However, in Fig. 3.66 (b) we have shown their magnitudes only. At a frequency f0, it is seen that the magnitude of XL is equal to XC as the two curves cut at point P. Since XL and XC are vectorially jXL and − jXc, the two reactances will cancel each other when frequency is f0. At f0 the impedance of the series R–L–C circuit is equal to R which is the minimum value of Z. In Fig. 3.66 (c), XL is represented as jXL and XC is represented as −jXC. The graph of X = XL − XC has also been drawn. The total impedance graph of Z shows that at f = f0, Z = R, i.e., at f0 the circuit offers minimum impedance, and hence maximum current will flow through the circuit.

At minimum value of Z, the current in the circuit will be maximum as I = V/R. This condition of the circuit when XL equals XC, Z = R, current is maximum and is called the resonant condition and the frequency, f0 at which resonance occurs is called the resonant frequency. At resonance, since XL equals XC, we can write

Alternately

or,

or,

or,

At resonance, frequency is f0, current I0 = V/R, power factor is unity, voltage drops across R, L, C are respectively, VR, VL, and VC and supply voltage V is equal to the voltage drop across the resistance VR.

Since at resonance, current is maximum and is very high, power dissipation I02R is maximum and the rate of energy storage in the inductor and the capacitor is maximum and they are equal. The value of R is usually small (this is the resistance of the inductive coil), and hence voltage drop across it, i.e., VR is also small as compared to the voltage drops across L and C. Voltage drops VL and VC are higher than VR. However, as VL = VC and they are in phase opposition as shown in Fig. 3.67 (b), the net voltage across L and C in series, VX is equal to zero. Thus, the supply voltage will be equal to VR.

Students will find it interesting to note that under the resonant condition the voltage across C or L will be many times more than the supply voltage. The power which is dissipated in the resistor is called active power. The energy which is stored in the inductor and the capacitor are due to reactive power. The energy stored in the inductor and the capacitor oscillates between them and the circuit as a whole appears to be a resistive only. The variation of circuit current as the frequency changes at different values of circuit resistance have been shown in Fig. 3.68 (a).

As can be noticed from the Fig. 3.68, at lower values of R, i.e., when R = R1 the sharpness of the current curve is increased. At the resonant frequency when current is at its maximum, the reactive power which oscillates between the inductor and the capacitor is much higher than the resistive power.

#### Quality factor

The ratio of the reactive power to the resistive power is called the quality factor. Quality factor is also defined as the ratio of voltage drop appearing across the inductor or the capacitor to the supply voltage. Thus,

Figure 3.67 Resonance condition in R–L–C series circuit: (a) circuit diagram; (b) phasor diagram

when

f = f0

or,

We had earlier calculated f0 as

Thus,

Figure 3.68 (a) Resonance curves for two values of resistance; (b) bandwidth of a resonant circuit

In terms of XC

Higher the ratio of reactive power to active power or higher the ratio of voltage across L and C to the supply voltage, the higher is the value of quality factor. Higher quality factor means sharper is the resonant factor curve and better is the ability of the network to accept current or power signals.

#### Bandwidth

Bandwidth is the range of frequencies for which the power delivered to the resistor is equal to half the power delivered to the resistor at resonance. As can be seen from Fig. 3.68 (b), the range of frequencies is (f2 − f1) and the corresponding current is i.e., 0.707 I0

Power delivered at resonance = I02 R

Half the power delivered at resonance =

= IBW2 R

Therefore, the range of frequencies within which current does not drop below 0.707 times the maximum value of current, i.e., I0 is called the bandwidth. See Fig. 3.68 (b). The frequencies f1 and f2 are often called the lower and upper cut-off frequencies. Bandwidth, BW = f2 − f1.

From Fig. 3.68 (b) it is seen that at both f1 and f2, the power delivered to R is equal to half the power delivered to R at resonance.

Using equation (3.18) we can write

or

or

R2 + (XL − XC)2 = 2R2

or

(XL − XC)2 = R2

or

or

From the above, the two values of frequencies, i.e., ω1 and ω2 are

Normally, is very small as compared to in a resonant circuit.

Therefore,

or,

Similarly,

ω2 − ω1 = ωBV

f2 and f1 are the higher and lower bandwidth frequencies. Let us now calculate the value of f0 / (f2 − f1)

Quality factor Q as we have seen earlier is

Thus,

From eq. (3.20), it is seen that if the quality factor is high, bandwidth will be narrow. The circuit will, therefore, allow only a narrow band of signal frequencies which are close to the resonant frequency. Such a circuit is, therefore, highly selective in allowing signals to pass through. High-quality-factor resonant circuits are also called ‘tuned circuits’, which will be studied in detail in electronic circuit design.

We have seen earlier that power dissipated at cutoff frequencies is half the power dissipated at resonant frequency. Hence, the cutoff frequencies f1 and f2 are also called half-power frequencies.

Example 3.33 An R–L–C series circuit has R = 10 Ω, L = 0.1 H, and C = 8 μf.

Calculate the following:

1. resonant frequency;
2. Q-factor of the circuit at resonance;
3. half-power frequencies and bandwidth.

Solution:

1. We know that the condition for the series resonance is
XL = XC

i.e.,

or,

Substituting values

2. Substituting values

3. Half-power frequencies are the frequencies corresponding to current which is 0.707 of the resonant current. They are f1 and f2, i.e., the lower, and upper frequencies forming the bandwidth.

and

Substituting

Bandwidth

= f2 − f1 = 18 Hz

Example 3.34 A circuit of R = 4 Ω, L = 0.5 H, and a variable capacitance C in series is connected across a 100 V, 50 Hz supply. Calculate:

1. the value of capacitance for which resonance will occur;
2. the voltage across the capacitor at resonance and the Q-factor of the circuit

Solution:

1. Applying condition for series resonance,
XC = XL

or,

= 2πf0L = 6.2 × 50 × 0.5 × 157 Ω

or,

2. Resonant Current,

Voltage across the capacitor,

Substituting values

= 3925 V

The students should note that a very high voltage of 3925 V is appearing across the capacitor while the supply voltage is only 100 V.

The Q-factor of the circuit =
= 39.25

Thus, the voltage multiplication across the capacitor as also across the inductor is 39.25 times at resonance.

Example 3.35 A resistor, a variable iron-core inductor, and a capacitor are connected across a 230 V, 50 Hz supply. By varying the position of the iron core inside the inductor coil, its inductance is changed. Maximum current of 1.5 A was obtained in the circuit by changing the inductance of the coil. At that time the voltage across the capacitor was measured as 600 V. Calculate the values of circuit parameters.

Solution:

We know that the maximum current flows at resonance when XL = XC and Z = R

Maximum current,

Therefore,

At resonance,

VL = VC = 600V
VL = Im XL = 600V

or,

2πfL = 400
XC = XL = 400

or,

or,

Example 3.36 An inductor, a variable capacitor, and a resistor are connected in series across a constant voltage, 100 Hz power supply. When the capacitor value is fixed at 100 μF, the current reaches its maximum value. Current gets reduced to half its maximum value when the capacitor value is 200 μF. Calculate the values of circuit parameters and the Q-factor of the circuit.

Solution:

Let resonant frequency be f0.

At resonance,

XL = XC

or,

or,

Substituting values,

= 25.3 × 10−3 H

Since XL = XC, the value of impedance at = R.

Maximum value of current,

At a frequency of 100 Hz, C = 200 × 10−6 F, current is reduced to half, i.e., impedance becomes equal to twice its value at resonance, i.e., equals 2R.

Impedance,

Current,

According to the problem

or,

or,

or,

(XL XC)2 = 3R2

or,

XL = 2πf0L = 628 × 25.3 × 10−3 = 15.88 Ω

Example 3.37 An inductive coil of resistance 10 Ω and inductance 20 mH are connected in series with a capacitor of 10 μF. Calculate the frequency at which the circuit will resonate. If a voltage of 50 V at resonant frequency was applied across the circuit, calculate the voltage across the circuit components and the Q-factor.

Solution:

R = 10 Ω, L = 20 × 10−3 H, C = 10 × 10−6 F

at resonance,

XL = XC

from which we get resonance frequency

= 356 Hz

At resonance, impedance of the circuit is equal to R. Therefore, the maximum current that will flow is equal to

To calculate the voltage drop across the circuit components, we calculate XL and XC at the resonance frequency first.

XL = 2πf0L = 2 × 3.14 × 356 × 20 × 10−3
= 44.7 Ω

voltage drop across L,

VL = I0 XL = 5 × 44.7 = 223.5 V

voltage drop across R,

VR = I0 R = 5 × 10 = 50 V

Note that the voltage drop across R is the same as the supply voltage, i.e., 50 V. Voltage drop across the capacitor should be the same as the voltage drop across inductive reactance XL. Let us calculate VC.

Substituting values

Thus,

VL = VC = 223.5 V

Q-factor

Example 3.38 A coil of inductance 1 mH and resistance 50 Ω connected in series with a capacitor is fed from a constant voltage, variable frequency supply source. If the maximum current of 5 A flows at a frequency of 50 Hz, calculate the value of C and the applied voltage.

Solution:

Resonant frequency,

or,

or,

= 0.0101F

At resonance,

XL = XC, Z = R

Voltage drop across R = Supply voltage = ImR = I0R

Thus, the applied voltage = 5 × 50 = 250V

Example 3.39 An inductive coil has a resistance of 2.5 Ω and an inductive reactance of 25 Ω. This coil is connected in series with a variable capacitance and a voltage of 200 V at 50 Hz is applied across the series circuit. Calculate the value of C at which the current in the circuit will be maximum. Also calculate the power factor, impedance, and current in the circuit under that condition.

Solution:

When current is maximum in an R, L, C series circuit, the circuit is under the resonance condition. At resonance, XL = XC and Z = R.

Here

XL = 25 Ω (given), f0 = 50 Hz

or,

At resonance, XL = XC, Z = R. The circuit behaves like a resistive circuit. Therefore, the power factor = 1.

Impedance, Z = R, and current is maximum.

Figure 3.69

#### 3.3.2 Resonance in AC Parallel Circuits

Let us consider an inductive coil and a capacitor in parallel connected across a constant voltage variable frequency supply source as shown in Fig. 3.69 (a). Practically, both the capacitor and the inductor will have some losses which should be represented by a small resistance in series. Here we assume the capacitor as a loss-less one while the inductor coil has some resistance which has been shown separately. The phasor diagram of voltage and current components have been shown in Fig. 3.69 (b). The line current I is equal to the in-phase component of IL with the voltage V, i.e., I = IL cos ϕL. At resonance, the current through the capacitor IC is balanced by IL sinϕL as shown. Thus, the reactive component of line current which is the phasor sum of IC and IL sin ϕL is zero. The condition for resonance is

IL sin ϕL = IC

or,

The condition of resonance is

XL XC =

To calculate resonance frequency, f0 we take, XL XC = = R2 +

or,

or,

or,

or,

or,

or,

This is the resonance frequency of a parallel L and C circuit.

If we consider the value of R as negligible, then the resonance frequency is

This value is the same as calculated for a series resonance circuit. The line current I is equal to IL cos ϕL which is the minimum current occurring at resonance. If the value of R is reduced, the cosine component of IL will get reduced. When R is made equal to zero, IL cos ϕL will be zero and the whole of IL will be reactive or wattless component and will be equal and opposite to IC.

I = IL cos ϕL is in phase with V when resonance occurs. The circuit impedance, Z0 is calculated as

From the condition of resonance

or,

Thus,

or,

Z0 is known as the equivalent impedance or ‘dynamic impedance’ of the parallel resonant circuit. It can be noticed that I = IL cos ϕL is in phase with the supply voltage. This shows that the circuit behaves like a resistive circuit only since the reactive component currents cancel each other. The impedance Z0 is therefore resistive only. Since current is minimum, impedance of the circuit, Z0 = L/CR is maximum under the resonant condition. Since current at resonance is minimum, a parallel resonant circuit is often referred to as a ‘rejector circuit’ meaning that a parallel resonant circuit tends to reject current at resonant frequency.

It may be noted that the current drawn from the supply at resonance, i.e., I = IL cos ϕL is minimum. The current circulating through the capacitor and the inductor, i.e., IC which is equal to IL sin ϕL is very high (IC >> or IL cos ϕL Since IC is many times more than I, we can say that parallel resonance is a case of current resonance. Here we recall that series resonance is a case of voltage resonance as voltage across the capacitor or the inductor is many times higher than the supply voltage.

#### Q-factor of parallel circuit

The ratio of circulating current between the two parallel branches, i.e., the capacitor and the inductor to the circuit line current is called the Q-factor or current magnification factor of the parallel circuit. Thus,

Value of

The effect of variation of frequency on circuit impedance and current has been shown in Fig. 3.70. At resonant frequency f0, the impedance is maximum, i.e., equal to The current at resonance is the minimum and is equal to I0 where

Figure 3.70 Variation of impedance and current in a parallel resonant circuit

Table 3.1 Comparison Between Series and Parallel Resonance

Current at resonance Maximum, I0 = V/R (Acceptor type) Minimum, I0 = (Rejector type)
Impedance at resonance Minimum, Z0 = R Maximum, Z0 = L/CR
Power factor at resonance Unity Unity
Resonant frequency
Magnification element Voltage Current
Magnification factor or Q-factor Q-factor Q-factor =

#### Bandwidth of parallel resonant circuit

Bandwidth of a parallel resonant circuit is determined the same way as in the case of the series resonant circuit. Bandwidth is the range of frequencies (f2 − f1 ) where the power dissipated is half of the power dissipated at the resonant frequency.

The critical parameters of series and parallel resonant circuits have been compared and shown in Table 3.1

We have seen that resonance in ac series and parallel circuits can take place at a particular frequency when a constant voltage, variable frequency supply source is applied across the circuit. The frequency at which resonance occurs are

for series circuit

and

for parallel circuit

If we neglect the small value of R, f0 for series and parallel resonance is the same.

If we have a constant frequency supply source, a resonance condition can also be achieved if we change the value of L or C creating a condition when

A resonance condition is created in tuning circuits of radio receiver sets by adjusting the values of circuit parameters.

Example 3.40 An inductive coil has a resistance of 10 Ω and inductance of 100 mH. This coil is connected in parallel with a capacitor of 20 μF. A variable frequency power at 100 V is applied across this parallel circuit. Calculate the frequency at which the circuit will reasonate. Also calculate the Q-factor, dynamic impedance of the circuit, and resonant current.

Solution:

Resonant frequency,

Substituting the given values,

= 112.48 Hz

Dynamic impedance,

Current at resonance,

Example 3.41 An inductive coil of resistance 5 Ω and inductive reactance 10 Ω is connected across a voltage of 230 V at 50 Hz. Calculate the value of the capacitor which when connected in parallel with the coil will bring down the magnitude of the circuit current to a minimum. Draw the phasor diagram.

Solution:

Figure 3.71

Before a capacitor is connected, current flowing through the inductor, IL is

cos ϕL = cos 64° = 0.438
sin ϕL = sin 64° = 0.895

If a capacitor is now connected in parallel, it must draw a current IC which will lead V by 90°. The magnitude of IC must be equal to IL sin ϕL so that these two currents cancel each other. In such a case, the resultant current, I is the in-phase current, i.e., IL cos ϕ].

IC = IL sin ϕL = 20.57 × 0.895 = 18.4 A

or,

= 254.7 μF

Magnitude of the in-phase current, i.e., the current which is in phase with the voltage, I = IL cos ϕ, is

I = IL cos ϕL = 20.57 × 0.438 9 A

This is the minimum current drawn by the circuit and is called the resonant current, I0.

Example 3.42 An inductor having a resistance of 4 Ω and an inductance of 20 mH are connected across a 230 V, 50 Hz supply. What value of capacitance should be connected in parallel to the inductor to produce a resonance condition? What will be the value of the resonant current?

Solution:

ZL = R + jωL, ω = 2πf = 2 × 3.14 × 50 = 314
ZL = 4 + j314 × 20 × 10−3 = 4 + j6.28 = 7.44∠57.5° Ω

For resonance, the current drawn by the capacitor in parallel must be equal to IL sin ϕL.

IC = IL sin ϕL = 30.9 × sin 57.5° = 30.9 × 0.843 = 26 A
= 360.2μF

Resonant current for parallel resonance is the minimum current which is the in-phase component, i.e., IL cos ϕ].

I0 = IL cos ϕL = 30.9 × cos 57.5° = 30.9 × 0.537
= 16.6 A

The phasor diagram representing the resonant condition has been shown in Fig. 3.72.

Figure 3.72

Example 3.43 Calculate the value of R1 in the circuit given in Fig. 3.73 such that the circuit will resonate.

Figure 3.73

Solution:

We know that at resonance the impedance of the circuit will be resistive only. We will calculate the value of impedance in a complex form and equate its imaginary part to zero to determine the value of R1.

Here, Z = R1 + j6 and Z2 = 10 − j4

During resonance, the imaginary part of Z will be zero.

Therfore,

j(60 − 4R1) (R1 + 10) − j2 (10R1 + 24) = 0

or,

60R1 − 4R12 + 600 − 40R1 − 20R1 −48 = 0

or,

4R12 = 552

or,

R12 = 138

or,

R12 = 11.74 Ω