Chapter 6
Power and Energy in Periodic Waveforms
CHAPTER OBJECTIVES
 To explain the need for sinusoidal waveforms and the importance of sinusoidal analysis.
 To explain the concepts of phase, phase difference, phase lag/lead, phase delay/advance, time delay/advance etc. in the context of sinusoidal waveforms.
 To introduce instantaneous power, cyclic average power and average power in periodic waveforms.
 To define effective value of periodic waveforms and illustrate rms calculations through examples.
 To develop and explain the power superposition principle and to emphasise the pitfalls in applying this principle.
 To develop an expression for effective value of composite periodic waveform and point out pitfalls in applying the result.
INTRODUCTION
Electrical power generation, transmission and distribution employ sinusoidal voltage and current waveforms to carry power. That, in itself, is a sufficient reason for a detailed discussion on steadystate analysis of circuits containing R, L and C.
A sinusoidal waveform is completely specified by three parameters. For example, if v_{S}(t) = A sin (ωt + θ) V this waveform is completely specified by three numbers – A, ω and θ. Therefore, sending a pure sine wave from a transmitter to a receiver in a communications context is pointless because such a waveform cannot carry any information other than that contained in just three numbers. And, for that matter, no waveform that is known completely beforehand can carry any information. Certain degree of uncertainty in the waveform to be transmitted is a precondition for information transmission from one location to another. Hence, the correct mathematical description of an informationbearing signal can only be a statistical description. However, despite this, the entire area of Electronic and Communication Engineering relies heavily on sinusoidal analysis of circuits and systems.
This raises two questions – (i) Why did the Electrical Power Industry prefer sinusoidal waveform to any other waveform? (ii) Why does Electronics and Communications Engineering concern itself with sinusoidal analysis though a singlefrequency sinusoid is hardly ever employed in a communication system?
These two questions are first answered in this chapter. Subsequently, the concepts of instantaneous power, average power, effective value of waveforms etc. are developed for sinusoidal waveforms as well as for other periodic waveforms.
6.1 WHY SINUSOIDS?
An Electrical Power System, like any other electrical circuit, contains passive electrical elements like resistors, inductors, capacitors, mutually coupled coils and active elements in the form of independent voltage sources and current sources. The elements in a power system are modelled by linear timeinvariant elements to a first degree of approximation. Resistors produce voltage drops across them that are proportional to the current flowing through them. Inductors demand a voltage that is proportional to the rate of change of current through them. Capacitors demand a voltage across them that are proportional to the integral of current through them.
Electrical power system is usually voltagedriven. That is, independent voltage sources connected at various points in the system serve as sources of power in the system. The voltage sources connected at various source nodes (i.e., the generating stations) drive currents through the series interconnections in the system, get modified by the voltage drops produced across various series path elements and appear as load voltage at load nodes in a modified form. Shunt elements connected from various nodes to the reference node in the system also influence this process of transformation of source voltages into load voltages.
Voltage drops across various elements thus modify the load voltage with respect to source voltage. These voltage drops are decided by a scaling of current by resistance value in the case of a resistor. It is decided by derivative of current in the case of an inductor and by integral of current in the case of a capacitor.
A timefunction retains its waveshape when multiplied by a constant. But, in general, it does not maintain its waveshape on differentiation and integration. Therefore, it follows that, in general, voltages and currents at various locations in an interconnected electrical network will have different waveshapes even if all sources in the network have same waveshape.
That would surely complicate things in an Electrical Power System. In fact, it will not be a viable system at all. That prompts us to raise the question – is there any waveshape that will be invariant to timedomain differentiation and integration?
A generalised exponential function, Ae^{αt}, has this property, as may be verified easily. The value of α can be complex. If α is real and positive, it represents a growing exponential. Such a waveform is not suited in an electrical system that is expected to operate steadily for extended duration. If α is real and negative, it represents a real decaying waveform that tapers down to zero sometime. An electrical system excited by a set of such sources will settle down ultimately to a state in which all voltages and currents everywhere will be zero. Obviously, such source waveforms cannot help the system to deliver power to loads in a steady manner for extended duration. If α = γ + jω, the exponential function is a complex function of time and is given by Ae^{γt} cos ω t + j Ae^{γt} sin ωt. We cannot generate an imaginary waveform in a physical system. But that problem can be solved by generating 0.5(Ae^{αt }+ Ae^{αt}) which is equal to Ae^{γt} cos (ω t). Here too, the same objections we raised for a real exponential will hold if γ is nonzero. Therefore, we conclude that 0.5(Ae^{jω t}+ Ae^{jω t}) = A cos ω t or –j0.5(Ae^{jω t}+ Ae^{jω t}) = A sin ω t are the possible choices for electrical power system source functions.
Thus, sinusoidal signals have gained their preeminent position in Electrical Power Engineering since they belong to a special class of timefunctions that preserve their waveshape under differentiation and integration. A linear network excited by sinusoidal sources of a particular frequency will have sinusoidal voltages and currents at the same frequency everywhere in the system in the long run.
We observe that a constant timefunction (i.e., DC voltages and currents) is a special case of Ae^{αt }with α = 0 and hence must satisfy the waveshapepreservation requirement. Thus, it must be possible to generate, transmit and deliver steady power to loads in an electrical system by means of DC sources. It is indeed possible and that was how Electrical Power Industry started in late 19th century. But the problem associated with DC system was total inflexibility with respect to voltage and current levels used in the system. For instance, if the customer had to be given 220V at his premises, the generators had to generate 220V and all the interconnection system had to work at that voltage level. As the load level increases, the current flow everywhere become excessive and generation/transmission become inefficient due to resistive losses everywhere in the system. It would have been very convenient if generation could be done at a voltage level economical from the point of view of electrical machine design and operation. Similarly, it would have been convenient if the transmission of power through transmission lines could be done at high voltage level so that the current level and consequently losses in lines would decrease. But this calls for generation at low voltage level, transmission at high voltage level and consumption at low voltage level. An efficient ‘voltage level conversion unit’ is needed for this. Such voltage level conversion equipment for DC at high power levels was simply not available at the initial stages of evolution of power systems in late 19th and early 20th centuries. And the same task turned out to be very easy in the case of a sinusoidal voltage system. A twowinding transformer can easily and efficiently change voltage levels in a system in the case of sinusoidal voltages. Thus, economic generation and transmission of huge quantities of electrical power became possible with sinusoidal voltage system and transformers. This is another reason why Electrical Power Industry is firmly committed to sinusoidal waveforms.
Advances in an area called Power Electronics resulted in power system loads becoming nonlinear in nature progressively from early 1980. Power Electronic Equipment (AC to DC converters, thyristorised DC motor drives, variable speed AC drives, uninterruptible power supplies, HVDC transmission systems etc.) process the sinusoidal system power using nonlinear devices for a variety of purposes. Nonlinear loads draw nonsinusoidal currents from sinusoidal voltages. Nonsinusoidal current waveforms, subjected to differentiation and integration in various circuit elements, result in nonsinusoidal voltage drops across them. These nonsinusoidal voltage drops, in combination with the sinusoidal voltages produced at various generating stations, result in a nonsinusoidal voltage at load nodes in a power system. This is called the Power System Harmonics Problem. Another very important feature of sine waves helps in analysing electrical systems that have nonsinusoidal periodic waveforms.
A broad class of periodic nonsinusoidal waveforms can be expressed as an infinite sum of sinusoidal waveforms with frequencies that are integer multiples of frequency of the periodic wave. For instance, a ± 1 V square wave with 1 cycle per second frequency can be expressed as (4/π)[sin2πt + (1/3) sin6πt + (1/5) sin10πt + ...+(1/n) sin 2nπt + ···]. Infinite terms are needed; but the amplitude of sine wave decreases as the order of the term increases. It is possible to truncate this kind of series expansion of a periodic waveform to obtain reasonably accurate results in circuit analysis problems. Thus, series expansion of nonsinusoidal periodic waveforms in terms of sinusoidal waveforms and Superposition Theorem will help us to solve a circuit driven by such periodic waveforms provided we know how to solve it for a sinusoidal waveform. Thus, even a power system under the influence of nonlinear loads can be analysed by sinusoidal analysis techniques with the help of this kind of series expansion for nonsinusoidal periodic waveforms.
Further, it turns out that, this series expansion of a periodic waveform leads to an expansion of even aperiodic waveforms in terms of sinusoidal waveforms as a limiting case. The series expansion of periodic waveform referred here is called Fourier Series, and the expansion of an arbitrary aperiodic waveform in terms of sinusoidal functions is called Fourier Transforms.
Fourier Series and Fourier Transforms along with Superposition Theorem help us to solve an electrical circuit driven by sources with arbitrary source functions by solving it for sinusoidal excitation. This is yet another factor that leads to preeminence of sinusoidal waveforms in circuit analysis. In fact, this is the reason why Electronics and Communication Engineers use sinusoidal analysis at all.
Thus, sinusoidal waveforms and sinusoidal analysis of circuits is of great importance to electrical and electronic circuits due to the following reasons:
 Sinusoidal waveforms preserve their waveshape in linear circuits.
 Sinusoidal waveform render the voltage levels in electrical and electronic systems flexible so that optimisation of subsystem performance in various parts of the system by choosing a suitable voltage level in that part of the system becomes possible. Transformers help us to realise this voltage flexibility.
 Periodic nonsinusoidal waveforms as well as a broad class of aperiodic waveforms can be expressed as a sum of sinusoids by Fourier series and Fourier transforms. Hence, circuit solution with such input waveforms can be obtained with relative ease if solution for a sinusoidal input is known.
6.2 THE SINUSOIDAL SOURCE FUNCTION
The sinusoidal voltage source function is generated in Synchronous Generators in Electrical Power Systems and by lowpower electronic oscillator circuits in Electronic Systems, Communication Systems and Instrumentation/Measurement Systems. A source will have to be switched on at some point in time. Switching on a source may be a simple affair of switching on the DC power supply as in the case of an electronic sinusoidal oscillator circuit or switching on the AC mains to a function generator in the laboratory. It may be a complicated affair involving a sequence of steps as in the case of a bringing a generator in a nuclear power station online.
Moreover, the sinusoidal voltage source may not start producing a sinusoidal output as soon as it is powered up. Usually, it goes through a transient period during which its output builds up. During this period, the output will not be a pure sine wave. After an initial period of adjustment, it starts delivering sinusoidal output to whatever that is connected at its output.
Even if it produces a sine wave right from the instant at which it is powered up, it may not start at zero position in the waveform or at peak position in the waveform.
We assume in this section that the sinusoidal sources have been powered up in the past and have become steady. Moreover, we assume that they started at zero position on a sine wave when they were switched on. The concepts we evolve are not really dependent on these assumptions. The assumptions are made only to render clarity to the discussion that follows.
6.2.1 Amplitude, Period, Cyclic Frequency, Angular Frequency
Consider a single sinusoidal voltage source that was powered up in the past. We start observing the waveform of the voltage output in an oscilloscope from a particular point in time. We assign zero value to the time variable at the instant we start our observation of the waveform. Thus, the source was powered up in the past with respect to the instant at which we start observing it. The observed waveform is shown in Fig. 6.21. The time variable t is used in the horizontal axis.
Fig. 6.21 Waveform of a sinusoidal voltage
The maximum positive value attained by the waveform is seen to be 10 V and the maximum negative value attained is –10 V. This quantity is called the amplitude of the sinusoidal waveform. It so happened that the waveform was crossing zero from negative value to positive value at t = 0. This zero crossing is called the positivegoing zero crossing. The zero crossing that happens when the voltage is crossing over from positive value to negative value is termed as negativegoing zero crossing. That t = 0 happens to be a positivegoing zero crossing is a coincidence. But as a result of that coincidence, we are now free to write the voltage waveform that we observe from t = 0 onwards as v(t) = 10 sinω t. We need to work out the meaning and value of ω.
We observe from Fig. 6.21 that the sinusoidal voltage completes one full cycle of variation in 20 ms. That is, if we start at any t and move through the waveform till we reach t + 20 ms, we will find that the instantaneous voltage at t + 20 ms is the same as the instantaneous voltage at t. Moreover, the shape of voltage variation in any (t + n × 20, t + 20 + n × 20) interval is same as in the interval (t, t+20), where the unit of time is in ms and n is a positive integer. Thus, the waveshape is repetitive with its basic repeating unit decided by any 20ms interval. That is, the waveform is periodic from the instant we start observing it. The period of this waveform is 20ms. In general, period of a periodic waveform is the time interval needed to complete one full cycle of the waveform. Symbol ‘T‘ is used to represent the period of a periodic waveform. In other words, it is the width of the basic repeating unit of the periodic waveform in the timeaxis.
The number cycles of variation that the waveform goes through in one second is defined as its cyclic frequency. The qualifier ‘cyclic’ is often dropped when there is no cause for ambiguity or when the unit employed makes it clear that it is cyclic frequency that is being referred to. The unit of cyclic frequency is ‘cyclespersecond’ and is given a name Hertz. Hertz is written in short form as Hz. The shortened form ‘cps’ is also used to designate the unit of cyclic frequency. The cyclic frequency of the waveform in Fig. 6.21 is 1/20ms = 50 Hz. Cyclic frequency is usually indicated by the symbol ‘f’.
A sinusoidal function of an angle is periodic with a period of 2π radians. Thus, the argument of the sinusoidal function in a sinusoidal waveform will go through an increment of 2π radians in one period. Therefore, the increment in the argument of the trigonometric function in one second will be 2π/Τ radians, where T is the period of waveform (= 1/f). This quantity, which represents the rate of change of angle argument of the sinusoidal function with respect to time, is defined as the angular frequency or radian frequency of the sinusoidal waveform and is usually represented by the symbol ω (lower case omega). The unit of angular frequency is radians/seconds, abbreviated as rad/s. Thus,
The sinusoidal waveform of voltage source v(t) = 10 sin 100πt shown against t in Fig. 6.21 is redrawn against the angular argument ωt in Fig. 6.22.
Fig. 6.22 Sinusoidal waveform v(t) plotted against ωt
A sinusoidal source voltage waveform that undergoes a positivegoing zero crossing at t = 0 can be expressed as v(t) = A sin ωt = A sin (2π/T) t = A sin 2πft V, where A is its amplitude, T is its period in s, f is its cyclic frequency in s^{1} (Hertz, Hz) and ω is its radian frequency or angular frequency in rad/s.
6.2.2 Phase of a Sinusoidal Waveform
Consider the observed sinusoidal waveform shown in Fig. 6.23. This waveform may be interpreted as a sinusoidal source powered up 17.5 ms prior to the start of observation. Assume that the source started delivering the sine wave output from zero value. The instantaneous value observed at the starting instant of observation is not zero, but –7.07 V. The amplitude is observed to be 10 V and the frequency is seen to be 50 Hz. Thus, we can express this waveform as v(t) = 10 sin (100πt +θ), where θ is an angle to be determined from the observed amplitude and instantaneous value at t = 0. The value at t = 0 is 10sinθ and this is seen to be –7.07 V. Therefore, θ = 45° = π/4 rad.
Fig. 6.23 Sinusoidal waveform with a nonzero phase
This waveform is plotted against ωt in Fig. 6.24. Note that the instantaneous value of v(t) at ωt = 100πt = π/4 rad is zero as it should be.
Fig. 6.24 A sinusoidal waveform with nonzero phase plotted against ωt
The argument of trigonometric function in the expression for a sinusoidal waveform is always in radians. However, it is permitted to write the angle θ in degrees provided the symbol of degree is clearly shown. That is, v(t) can be written as 10 sin (100πt – 45°) V, but not as 10 sin(100πt – 45). In the latter case, 45 will be interpreted with radian unit. Moreover, though the form 10 sin (100πt – 45°) V is permitted, 45° has to be converted into radians before subtracting from the value of 100πt for some particular t before evaluating the sine function. In short, the form 10 sin (100πt – 45°) is allowed only as a notation and not for function evaluation. The angular argument has to be in radians when an evaluation of trigonometric function is attempted.
The quantity θ in v(t) = A sin (ωt + θ) is defined as the phase of the sinusoidal function.
6.2.3 Phase Difference Between Two Sinusoids
Consider the situation in Fig. 6.25. Two observers – A and B – use XYRecorders A and B, respectively to record the output from two sinusoidal voltage sources v_{1}(t) and v_{2}(t) which had been powered up in the past. A closes the two switches onto his recorder at t = 0. t represents the timeaxis chosen by him. B closes the switches onto his recorder at t' = 0. t' represents the timeaxis chosen by B. v_{1}(t) has an amplitude of V_{m1} and v_{2}(t) has an amplitude of V_{m2}.
Fig. 6.25 Simultaneous observation of two sinusoidal sources by two observers with different starting instants for observation
The waveforms recorded by Observer A are shown in Fig. 6.26(a) by the solid curve. The dotted curve shows the sinusoidal variation of sources prior to recording and will not show up in the recorder output. The waveforms in Fig.(a) are normalised with respect to their respective amplitude values to obtain the waveforms in Fig.(b).
Fig. 6.26 Waveform observation by Observer A
Two pairs of similarly located waveform points within a cycle period are located in the waveshape of v_{1}(t)/V_{m1} and v_{2}(t)/V_{m2} as shown in Fig. 6.26(b). Observer A notes that similarly located points in the two waveforms are separated by 45° in angle argument. A also notes that points on v_{2}(t) come after (in a visual sense) similarly located points on v_{2}(t).
Similar observations recorded by Observer B in ωt' axis is shown in Fig. 6.27.
Fig. 6.27 Waveform observations by Observer B
B too measures 45° angular separation between similarly located points on normalised v_{1}(t) and v_{2}(t). Moreover, B too observes that points on v_{2}(t) come after similarly located points on v_{1}(t).
The angular difference between similarly located points within a cycle period on two normalised sinusoidal waveforms (normalised with respect to their respective amplitude values) with same frequency is defined as the phase difference between them. The phase difference between two sinusoids is independent of choice of origin in t or ωt axis. The precedence relationship [i.e., which comes after (in a visual sense) which] between them in t or ωt axis too is independent of choice of origin.
However, based on the observed amplitudes, values at origin and the position of first zerocrossing, A will write the sinusoidal functions as v_{1}(t) = 10 sin100πt V and v_{2}(t) = 5 sin(100πt – 45°) V. Moreover, B will conclude that v_{1}(t') = 10 sin(100πt'  60°)V and v_{2}(t') =5 sin(100πt' – 105°)V Thus, the phase of v_{1}(t) is 0° and phase of v_{2}(t) is –45° as far as A is concerned. And they are –60° and –105°, respectively as from B’s point of view.
The phase of a sinusoidal waveform depends on the choice of origin in t or ωt axis. Phase difference between two sinusoidal waveforms at same frequency does not.
When a waveform point on a sinusoidal function v_{2}(t) appears after a similarly located point on the waveform of another sinusoidal function v_{1}(t) with same frequency, v_{2}(t) is said to lag v_{1}(t) in phase and the corresponding phase difference between them is called a lag phase angle under this condition.
Similarly, when a waveform point on a sinusoidal function v_{2}(t) appears before a similarly located point on the waveform of another sinusoidal function v_{1}(t) with same frequency, v_{2}(t) is said to lead v_{1}(t) in phase and the corresponding phase difference between them is called a lead phase angle under this condition.
It must be obvious that if v_{2}(t) lags v_{1}(t), then v_{1}(t) must necessarily lead v_{2}(t). Moreover, if v_{2}(t) lags v_{1}(t), then v_{2}(t') will also lag v_{1}(t'), where t' is a new time variable as result of a different choice of origin.
6.2.4 Lag or Lead?
Similarly located points on two sinusoidal waveforms with same frequency have to be located within a period of the waveforms. But this leads to two choices for locating the point on the second waveform after having chosen a point on the first waveform (refer Fig. 6.28).
Fig. 6.28 Relationship between phase lag and phase lead
We locate point B on normalised v_{1}(t) first. We are free to locate the similarly located point on normalised v_{2}(t) on either side of B within a span of 2π radians or 360°. This gives us two choices – point C and point D on the second waveform. If we choose point C, we can conclude that v_{2}(t) leads v_{1}(t) by 315°. If we choose point D, we conclude that v_{2}(t) lags v_{1}(t) by 45°. Therefore, a lag angle of θ radians and a lead angle of (2π–θ) radians mean the same. As a convention, we favour the angle that turns out to be less than 180° (or π radians). Thus, in Fig. 6.28, we will term it as a lag angle of 45°.
6.2.5 Phase Lag/Lead Versus Time Delay/Advance
It leaves a certain ambiguity about which waveform is after the other. If we accept the point C on v_{2}(t) in Fig. 6.28 as the point corresponding to point B on v_{1}(t), we conclude that v_{2}(t) comes before v_{1}(t). Similarly, If we accept point D on v_{2}(t) as the point corresponding to point B on v_{1}(t), we conclude that v_{2}(t) comes after v_{1}(t). However, note carefully that we had been careful to keep the precedence relationships –before and after – only in relation to our visual perception of the waveform plots. We have not yet ascribed temporal significance to these terms. That is, we have not stated till now that if a waveform v_{2}(t) comes after v_{1}(t) in a visual sense, then v_{2}(t) started later than v_{1}(t) in time. In other words, we have not correlated the phase difference between two waveforms with time delay or time advance between them. The term ‘phase lag’ tends to give us an impression that the waveform that lags behind suffered some time delay with respect to the other waveform. But this impression can be wrong. Similarly, the waveform that leads ahead of another waveform did not necessarily start earlier. The reader is cautioned against equating a ‘phase lag’ with a ‘time delay’ and a ‘phase lead’ with a ‘time advance’ indiscriminately. There are situations in which a ‘phase lag (lead)’ implies a ‘time delay (advance)’ – in that case, we will term the phase lag (lead) as ‘phase delay (advance)’. And there are situations in which lag/lead cannot be uniquely correlated to delay/advance in timedomain.
Consider the two waveforms v_{1}(t) and v_{2}(t) in Fig. 6.29(a). Additional information in the form of dotted curves is also shown in Fig.(a). The frequency of waveform is 50 Hz. The waveforms in Fig.(a) show that the source v_{1}(t) started generating a sinusoidal voltage at 20 ms before the observation started and v_{2}(t) started only 2.5 ms later. It is also clear that v_{2}(t) lags v_{1}(t) by 45°. Thus, a time delay of 2.5 ms has resulted in a phase lag of 45°. Since we know from the additional information provided in the form of dotted curves that the phase difference between two sources resulted from a time delay, we can term this 45° phase lag as a 45° phase delay too. Obviously, the following relation between time delay and phase delay holds:
Phase delay ɸ in radians = ω × time delay t_{d }in seconds
i.e., ɸ = ωt_{d}
Similar statement can be arrived at in the case of time advance too, provided we know from information other than we obtained from observing the two waveforms from t = 0 that the observed phase difference is due to a time advance.
phase advance ɸ in radius = ω × time advance t_{d} in seconds
i.e., ɸ = ωt_{d}
Now, consider the waveforms in Fig. 6.29(b). Here, v_{1}(t) started 17.5 ms earlier than v_{2}(t). But if we go only by the observation from t = 0 onwards, we will conclude that v_{2}(t) leads v_{1}(t) by 45° or equivalently v_{2}(t) lags v_{1}(t) by 315°, and, as per the agreed convention, we will settle for ‘v_{2}(t) leads v_{1}(t) by 45°’. But, in the light of the additional information given in the form of dotted curves, a translation to the effect that v_{2}(t) started 2.5 ms earlier than v_{1}(t) will be in error. Actually, v_{2}(t) started 17.5 ms (315°) after v_{1}(t) and hence phase delay of v_{2}(t) is 315° and time delay of v_{2}(t) is 17.5 ms with respect to v_{1}(t). The conclusion from observation from t = 0 onwards can also be stated as v_{1}(t) lags v_{2}(t) by 45°. Again, a translation to the effect that v_{1}(t) started 2.5 ms after v_{2}(t) started is wrong in the light of additional information given. Actually, v_{1}(t) started 17.5 ms before v_{2}(t) started and hence v_{1}(t) has a phase advance of 315° and a time advance of 17.5 ms with respect to v_{2}(t).
Fig. 6.29 Illustrating phase delay and time delay
Thus, additional information is required to translate an observed phase lag/lead relationship between two sinusoidal waveforms into a phase delay/advance (equivalently, time delay/advance) relationship between them. Phase lag is not necessarily a phase delay and phase lead is not necessarily a phase advance. Phase lag does not necessarily imply time delay and phase lead does not necessarily imply time advance.
The additional information needed to decide time delay/advance from phase lag/lead is not usually available in the case of multiple sinusoidal source waveforms in a complex electrical system. But then, we do not usually need the time delay/advance information in Electrical Power Systems.
There is one situation in which this additional information needed is invariably available. Consider a situation in which a sinusoidal voltage source is applied to a linear electrical network at t = 0. The circuit variables respond to this excitation and assume pure sinusoidal variation at the same frequency as that of sinusoidal excitation in the long run. There will be a definite phase difference between a response variable (may be a current in some element or a voltage across some element) and the source function. No physical system can produce a response before the excitation is applied to it. Response always follows the excitation in a physical system and cannot precede excitation. This intuitively obvious fact is known as the ‘law of causality’ for physical systems. Thus, law of causality of physical systems effectively states that the response will be delayed with respect to excitation. Therefore, the response sinusoid in an electrical circuit will always be delayed with respect to the excitation sinusoid quite regardless of whether the phase difference is a lag angle or lead angle. A phase lead that the response variable exhibits with respect to excitation variable in a physical electrical network has to be understood as a phase delay that is more than π radians and a time delay that is more than halfperiod. The reader is cautioned against the commonly made mistake of assuming that the apparent phase lead exhibited by a response variable implies a phase advance or time advance.
Example: 6.21
Two sinusoidal waveforms, x(t) and y(t), recorded from t = 0 are shown in Fig. 6.210. x(0) = –2.571 and y(0) = 1.25 (i) Express x(t) and y(t) as sine functions and identify their amplitude, period, cyclic frequency, radian frequency, phase and phase difference. (ii) If no additional information is available, list all possible time delay/advance relations between the two assuming that both started at positivegoing zerocrossing position. (iii) If x(t) is voltage waveform that was powered up and applied to a linear electrical circuit long back in the past and y(t) is the current flow in some element in that circuit, find the time delay/advance between the two and phase delay/advance between the two.
Fig. 6.210 Sinusoidal waveforms for Example 6.21
Solution
 The period of both waveforms, T = 20 ms. Therefore, f = 50 Hz and ω = 100π rad/s. Amplitude of x(t) is 4 and amplitude of y(t) is 2.5.
Therefore, x(t) = 4 sin (100πt + θ_{χ}) and y(t) = 2.5 sin (100πt + θ_{γ}), where θ_{x} and θ_{y} are the phases of x(t) and y(t), respectively. The values of x(t) and y(t) at t = 0 are given as –2.571 and 1.25, respectively.
∴sin θ_{x} = –2.571/4 = –0.6428 ⇒ θ_{x } = – 40° or –140°.
The choice between –40° and –140° is made by observing that if it is –40°, the first zerocrossing of the waveform that takes place at ωt = –θ_{χ} will take place within the first quarter cycle and if it is –140°, the first zerocrossing after t = 0 will take place after first quarter cycle. In the present case, x(t) crosses zero within first 5 ms (first quarter cycle) and hence θ_{χ} = –40°.
∴x(t) = 4 sin(100πt – 40°)
Similarly, sin θ_{y} = 1.25/2.5 = 0.5 ⇒ θ_{y} = 30° or 150°. Now the first zero crossing will take place at ωt = 150° position (i.e., in the second quarter cycle) if θ_{y} = 30° and it will take place at ωt = 30° position (i.e., in the first quarter cycle) if θ_{y} = 150°. Since it takes place within the second quarter cycle in the present case, θ_{y} = 30°
∴y(t) = 2.5 sin(100π t + 30°)
The phase difference between the two waveforms has a magnitude of 30°–(–40°) = 70° and x(t) lags y(t) by 70°. Equivalently, y(t) leads x(t) by 70°. Equivalently, x(t) leads y(t) by 290° and y(t) lags x(t) by 290°.

70° phase difference translates to 20 × 70/360 ≈ 3.89 ms time interval. The following are the possibilities:
 y(t) started (3.89+ 20n) ms before x(t) and therefore y(t) has a phase advance of (70°+ n360°) with respect to x(t), where n = 0, 1, 2, 3, ... . This may also be restated as x(t) has a phase delay of (70°+ n360°) with respect to y(t).
 x(t) started (16.11+ 20n) ms before y(t) and therefore x(t) has a phase advance of (290°+ n360°) with respect to y(t) where n = 1, 2, 3, . . This may be restated as y(t) has a phase delay of (290°+ n360°) with respect to x(t).
 Now, x(t) is the cause and y(t) is the effect in a physical electrical circuit. Therefore, by law of causality, y(t) can have only a phase delay with respect to x(t). Therefore, y(t) has a phase delay of (290°+ n360°) with respect to x(t).
However, this does not mean that the circuit waited for (16.11+ 20n) ms after the voltage waveform was applied to it doing nothing in that interval and then started producing a sinusoidal current in the element under consideration! What happens in the electrical circuit is that, as soon as the voltage waveform is applied, the circuit starts a mixed response that includes even nonsinusoidal terms caused by the electrical inertia of the circuit. This period is called the transient period. The nonsinusoidal components in the response die down to zero as time progresses. After a sufficiently long duration decided by circuit parameters, a steadystate comes up in the circuit in which all the response variables become pure sinusoidal waveforms at the same frequency as that of excitation. And, by that time, the current y(t) would have acquired a steady phase delay of (290°+ n360°) with respect to the applied voltage x(t). The value of n that is applicable can be obtained only if the circuit is known in detail and a full circuit solution is obtained.
Phase lag/lead between various voltages and currents in an Electrical Power System has profound implications in the economic operation of the system. Phase delay and time delay between various sinusoidal waveforms in an electronic system or communication system has great significance in terms of waveform distortion and loss of information contained in a waveform. Therefore, Electrical Power Engineers pay a great deal of attention to phase lag/lead between waveforms whereas Electronics and Communication Engineers place even higher emphasis on phase delay/advance and time delay/ advance between waveforms. That is why these terms were discussed in detail in this section.
6.3 INSTANTANEOUS POWER IN PERIODIC WAVEFORMS
A waveform v(t) is said to be periodic with a periodicity of T s if v(t+nT) = v(t) for all integer values of n and for all t. This implies that it must be possible to identify a basic section of the waveform that lasts for T s and that repeats to infinite extent into the past and into the future. Thus, a waveform is strictly periodic only if it is everexistent. But, in practice, waveforms are switched on at some definite time instant. Such switched waveforms cannot be called periodic waveforms in the strict sense of definition of periodicity. However, we can view them as periodic waveforms for circuit analysis purposes provided we focus our attention to time instants located far away from the instant at which the waveform was switched on.
Fig. 6.31 A twoterminal element with voltage and current marked as per passive sign convention
Consider a twoterminal electrical element with the current and voltage variables marked as per passive sign convention in Fig. 6.31.
The voltage difference v_{AB} between two points A and B is the work to be done in moving +1 C of charge from B to A. Energy has to be spent in carrying charge from a lower potential point to higher potential point. Similarly, energy is released when a charge is allowed to fall through a higher potential point to lower potential point. The amount of charge that went through the element from a higher potential point to lower potential point in one second is given by i(t). Therefore, the product of v(t) and i(t) must be the energy released into element in one second. The rate of change of energy is defined as instantaneous power and is denoted by p(t).
Therefore, instantaneous power is delivered to a twoterminal element, p(t) = v(t) i(t), where v(t) and i(t) are the element variables defined as per passive sign convention.
Then, energy delivered to a twoterminal element is obviously given by
where E(0) is the total energy dissipated in the element from infinite past to t = 0.
And the relation between the energy function E(t) and the instantaneous power p(t) is given by Let E_{i}(t) be the energy dissipation function (i.e., the net energy delivered from –∞ to t) of the i^{th} element in a b element electrical circuit. Then the total energy dissipation function of the circuit is The circuit considered as a whole is an isolated system and the total energy in an isolated system is a constant by Law of Conservation of Energy. Therefore,
This implies that if some elements of the circuit are receiving energy, then some other elements must be losing energy.
Since the statement of conservation of energy is true on an instanttoinstant basis, we can differentiate both sides of the above equation with respect to time:
But each term of the type can be interpreted as the instantaneous power delivered to the i^{th} circuit element.
Therefore, sum of instantaneous power delivered to all elements in a circuit will be zero. Equivalently, sum of instantaneous power delivered by all elements in a circuit will be zero. This implies that in any circuit some elements will be delivering positive power and the remaining elements will be receiving positive power at all instants of time. The sum of positive powers delivered will be equal to the sum of positive powers received (or consumed) at all t. This is the principle of conservation of instantaneous power which is valid for all isolated circuits containing arbitrary kind of elements. Isolated circuit is one that has no interaction with environment.
Principle of Conservation of Instantaneous Power – The sum of instantaneous power delivered to all elements in an isolated circuit will be zero, i.e., p_{1}(t) + p_{2}(t)......+p_{b}(t)=0 where b is the total number of elements in the circuit.
Example: 6.31
A DC voltage of 10 V is switched on to a 10Ω resistor at t = 0. Find and plot p(t) and E(t) for t ≥ 0.
Solution
The waveforms of applied instantaneous power and the energy dissipated are shown Fig. 6.32.
Fig. 6.32 Instantaneous power and energy in Example 6.31
Example: 6.32
A symmetric square wave of 100 Hz frequency and ±10V amplitude is switched on to a resistor of 10Ω at t = 0 at the positivegoing zero crossing of the square wave. Find and plot the instantaneous power and energy dissipation in the resistor.
Solution
When a symmetric square wave is squared, it becomes a constant quite irrespective of its frequency. Therefore, [v(t)]^{2} = 100 V^{2} for t ≥0.
∴ p(t) = 10 W for t ≥0 and E(t) = 10 t J. The plots are same as in Fig. 6.31.
Therefore, we conclude that applying a symmetric square wave voltage of amplitude V and frequency f to a resistive load has the same effect as applying a DC voltage of V to it as far as the instantaneous power and energy dissipation function are concerned.
Example: 6.33
A periodic rectangular pulse voltage shown in Fig. 6.33 is applied to a resistor of 10 Ω. Find and plot instantaneous power and energy dissipation function.
Fig. 6.33 The applied voltage waveform for Example 6.33
Solution
The waveform of p(t) is obtained by squaring the waveform of v(t) and dividing by 10. E(t) is obtained by integrating (i.e., evaluating the area under p(t) curve) the power waveform. The two plots are shown in Fig. 6.34 for the time range [0ms, 30ms].
Energy delivery to the load takes place through power pulses in this example. As a consequence, the energy dissipated in the resistor rises nonuniformly with time. However, E(t) is either increasing or constant at all t consistent with the fact that p(t) never goes negative. The plot of E(t) versus t shows a dotted line. This line is obtained by joining the points representing the total energy dissipation in the resistor at the end of a cycle. It turns out to be straight line with a slope of 0.01 J/ms = 10 W. The power pulse located within a cycle of p(t) has an amplitude of 50 W and duration of 1 ms. The cycle duration is 5ms. Hence, the energy delivered to the resistor in each cycle is 50 W × 0.001 s = 0.05 J. The same energy would have been delivered to the resistor in 5ms if a constant power of 10W were available for the entire cycle. We note that 10 W is the average of instantaneous power over one cycle period.
Fig. 6.34 Instantaneous power and energy plots for Example 6.33
Therefore, a line joining the endofcycle energy values in the energy versus time diagram will have a slope that is equal to the average value of instantaneous power over a cycle.
Example: 6.34
v(t) = 10√2 sin t V is applied to a 10 Ω resistor from t = 0 onwards. Plot p(t) and E(t) for t ≥ 0.
The waveform of p(t) is obtained by squaring the waveform of v(t) and dividing by 10. E(t) is obtained by integrating (i.e., evaluating the area under p(t) curve) the power waveform. Therefore,
Therefore, the instantaneous power has a frequency that is double that of the frequency of voltage. Moreover, p(t) contains a DC component. The plots of p(t) and E(t) are shown in Fig. 6.35 for 0 ≤ t ≤ 4πs.
The cycle period of power waveform is πs. The values of E(t) at integer multiples of cycle period of power waveform are joined by a dotted line as shown in Fig. 6.35. This line turns out to be a straight line with a slope of 10 W. The average value of p(t) over one cycle of πs is also 10 W, as shown in the following derivation:
The energy function in this example too is a monotonically increasing function of time. Example 6.35 considers a situation in which the energy function is not monotonic on t.
Example: 6.35
A sinusoidal voltage source v(t) =10√2 sin t V is connected across a sinusoidal current source i(t) = √2 sin(t – 60°) A at t = 0. i(t) flows out of the positive terminal of v(t). Find the instantaneous power delivered by the voltage source to the current source and the corresponding energy function and plot them.
Fig. 6.35 Plots of instantaneous power and energy in Example 6.34
Solution
Let p(t) be the instantaneous power delivered by the voltage source to the current source. Then,
Energy is obtained by integrating this function over t = 0 to t = t.
The waveforms of voltage, current, power and the plot of energy function are shown in Fig. 6.36.
Voltage and current have a frequency of 1 Hz whereas the power waveform has a frequency of 2 Hz. Power waveform is bipolar. The voltage source delivers energy to the current source for some time during a cycle and the current source delivers energy to the voltage source for the remaining time. However, the net energy delivered in one cycle is from voltage source to current source. The energy plot shows that E(t) is nonmonotonic on t. This must be so since the power waveform is bipolar.
The line joining the energy values at the end of cycles of power is seen to be a straight line with a slope of 5 W. The average value of p(t) over a cycle must then be 5 W. It is verified to be 5 W in the derivation that follow.
Fig. 6.36 Waveforms and plots for Example 6.35. Upper traces show voltage, current and power.
Example: 6.36
An electrical element draws a current of i(t) = –√2 cos 100πt A from a sinusoidal voltage source of v(t) = 10√2 sin 100πt V. Find and plot instantaneous power delivered to the load and energy delivered to it as functions of time.
Solution
p(t) = –20sin 100πt cos 100πt = –10 sin 200πt W and E(t) = –0.05(1 – cos 200πt) J.
Power waveform has 100 Hz frequency whereas the voltage and current waveforms have 50 Hz frequency. The period of the power waveform is 10 ms.
Average value of power over a cycle Therefore the energy function cannot be a growing function. The net change in E(t) over a cycle of power waveform will be zero. It has to be an alternating function of time with a possible average content, indicating that there is no net energy transfer from source to load over a cycle of power waveform. The plots shown in Fig. 6.37 confirm this.
Fig. 6.37 Waveforms and plots for Example 6.36
6.4 AVERAGE POWER IN PERIODIC WAVEFORMS
The following points emerge from the discussion on instantaneous power in Section 6.3.
 Instantaneous power delivered to an element is a nonconstant function of time in general.
 If the voltage across the element and current through the element are periodic waveforms with period T and zero average value over a cycle period, the instantaneous power will be a periodic waveform with period 0.5T and may have a nonzero average value over its cycle period of 0.5T.
 The energy delivered to the element will also be a function of time. The value of total energy delivered to the element at endofcycle points will fall on a straight line with a slope equal to the average value of instantaneous power waveform over its cycle period.
 If the instantaneous power waveform is unipolar, the energy function will be monotonic on t. If instantaneous power waveform is bipolar, the energy function will be nonmonotonic.
A 100 W incandescent lamp draws about 0.615 A peak sinusoidal current when a 325 V peak sinusoidal voltage at 50 Hz is applied across it. This results in a p(t) = 200 sin^{2}200πt W of instantaneous power in it. p(t) can also be expressed as p(t) = (100 – 100 cos 200πt) W. Thus, the instantaneous power varies from 0 to 200 W and goes back to zero in 10 ms. Does the lamp filament respond to this power input variation?
It does respond to the power variation. However, 10ms is too small a time interval for significant variations in the temperature of the filament to take place. This is due to the thermal capacity of the lamp system that tends to behave like thermal inertia when it comes to changes in lamp temperature. Thus, though the lamp responds to both the 100 W constant component and the 100 W cosine term in instantaneous power, the response of temperature variable to the 100 W cosine term is very small compared to the response to the 100 W constant term. Thus, the lamp temperature and hence its light output is more or less constant in the long run (after about 50 to 100ms of switching on in this context) corresponding to the 100 W constant term in p(t), with the 100 W rippling term contributing only a negligible amplitude oscillation in them. And, persistence of human vision virtually blots out even this small rippling component in light output. Hence, we do not get to see the small amount light flicker that takes place invariably due to the power changing periodically with a period of 10 ms. If we apply a 5 Hz voltage instead of 50 Hz voltage to a lamp, the oscillating component in lamp light output will be much more in amplitude and the frequency of this oscillating component will be 10 Hz. We will experience a prominent flicker in the lamp light output.
Thus, we conclude that, if the frequency of pulsation in instantaneous power delivered to an electrical load (lamps, heating element, motors etc.) is sufficiently high, the output from the load (light, temperature, torque, speed etc.) will be constant in the long run. The magnitude of this constant output is decided by the average of instantaneous power over its cycle period. The load will ignore the pure alternating component/s in instantaneous power. Therefore, the average value of p(t), averaged over its cycle period, is a much more relevant quantity in practice.
The Cycle Average Power in the context of periodic waveforms is defined as the cycle average of instantaneous power over a cycle of instantaneous power and is denoted by P, i.e., where T is the period of the periodic voltage and current waveforms and t is any arbitrary time instant after t = 0. p(t) is assumed to be zero for t < 0. In practice, the integration is carried out from the beginning of a power cycle to the end of that cycle.
The Average Power contained in an instantaneous power waveform and the Cycle Average Power are two different concepts altogether. First of all, there can be a cycle average power only if p(t) is a periodic waveform. But average power can be defined and calculated for any p(t). Let v(t) be the voltage across an element and i(t) be the current through it over an interval of time denoted by [t_{1}, t_{2}]. Then the energy delivered to the element during this interval is given by the area under p(t) = v(t) i(t) from t_{1} to t_{2}. Then, the Average Power (P_{av}) delivered during this interval is the value of constant power that would have delivered the same amount of energy to the element in the interval between t_{1} to t_{2}.
The instant t_{1} is usually chosen to be the instant at which the voltage was applied to the element and the instant t_{2} is the instant at which the supply to the electrical element was switched off. If the two instants are chosen this way, the P value gives the average rate at which energy was delivered to the load element during the entire period of connection.
A simple relation exists between Average Power (P_{av}) and Cyclic Average Power (P) in the context of periodic voltages and currents. Let v(t) = V_{m} sin ωt V and i(t) = I_{m} sin (ωt + θ) A. Let the interval duration t_{2}– t_{1} extend over a large number of cycles of p(t). The instantaneous power in this case will be periodic with a period of 0.5T. But t_{2} – t_{1} may not be an integer multiple of 0.5T. Hence, we express t_{2} – t_{1} as 0.5nT + 0.5kT where k is a real number between –1 and 1 and n is an integer.
We used t_{2} – t_{1} = 0.5nT + 0.5kT in the last step. sin (πn + πk) is zero if there are integer periods of length 0.5T within t_{2} – t_{1}. If not, the magnitude of second term in the expression for P_{av} is upper bounded by V_{m}I_{m}/2π(n+k). For a sufficiently large n, i.e., if the length of the interval over which the average power is calculated is very large compared to the period of instantaneous power waveform, then, the second term becomes negligible. The first term is same as cycle average power. Therefore, Average Power, P_{av} = Cycle Average Power, P, if the waveforms last for sufficient duration compared to their period. If v(t) and i(t) persist for more than 20 or more cycles, the error in taking P as P_{av} will be < 1%.
Example: 6.41
The periodic 50Hz voltage waveform applied across a 5 Ω resistor is shown in Fig. 6.41. (i) Find the average power delivered to the load. (ii) Find the DC voltage that will deliver the same average power. (iii) Find the amplitude of a 50 Hz sinusoidal voltage that will deliver the same average power.
Fig. 6.41 Applied voltage waveform in Example 6.41
Solution
 The power waveform will be described by the following equation for the first period:
Energy delivered in one cycle of p(t) = The area under p(t) over one cycle (10ms) of p(t) = 2 × area under p(t) for 0 to 5ms interval =
Therefore, Cycle Average Power = 6.67 J/0.01 s = 667 W
Therefore, P_{av} = 667 W
 The required DC voltage V is such that V^{2} /5 = 667. Therefore, V = 57.74 V
 Let v(t) = V_{m} sin 100πt. Then, i(t) = 0.2 V_{m} sin 100πt and p(t) = 0.2 V_{m}^{2} sin^{2}ωt W. Therefore, cycle average power = 0.2 V_{m}^{2} ÷2 = 0.1 V_{m}^{2}. This has to be 667 W. Therefore, the required amplitude of sinusoidal voltage is V_{m} = 81.67 V
Example: 6.42
A 12V battery delivers power to a power electronic system through its internal resistance of 0.05Ω as shown in Fig. 6.42. The figure also shows that the current drawn by the load is a pulsed current with period of T = 1 ms. The load draws the current in such a way that the average value of current is kept constant at 10 A while d and I vary under different operating conditions. Calculate the average power delivered by the source, average power delivered to load and efficiency for (i) I = 100 A (ii) I = 40 A, (iii) I = 12 A. and (iv) I = 10 A.
Fig. 6.42 Circuit and waveform forExample 6.42
Solution
Since the average value of i(t) is kept constant, Id must be constant.
Average power delivered by the source is This value is independent of I and d as long as the average value of current is kept at 10 A. The power delivered by a constant voltage source is equal to product of source voltage and the cycle average value of current.
Average power dissipated in the internal resistance, P_{R}, is
We have used the fact that Id = 10 A in the last step.
 I = 100 A. Therefore, d = 10/100 = 0.1.
P_{R} = 0.5 × 100 = 50 W., P_{S} = 100 W, ∴Load power, P_{L} = 100 – 50 = 50 W
Efficiency = 50%
 I = 40 A. Therefore, d = 10/40 = 0.25
P_{R} = 0.5 × 40 = 20 W. P_{S} = 100 W, ∴Load power, P_{L} = 100 – 20 = 80 W
Efficiency = 80%

I = 12 A. Therefore, d = 10/12 = 0.833
P_{R} = 0.5 × 12 = 6 W. P_{S} = 100 W, ∴Load power, P_{L} = 100 – 6 = 94 W
Efficiency = 94%

I = 10 A. Therefore, d = 10/10 = 1, i(t) becomes a constant current of 10 A.
P_{R} = 0.5 × 10 = 5 W. P_{S} = 100 W,
∴Load power, P_{L} = 100 – 5 = 95 W
Efficiency = 95%
Example: 6.43
A DC voltage source of V V delivers a current i(t) to an external load through its internal resistance of R Ω. Show that, among the infinite possible periodic waveforms for i(t) with the same value of cycle average, i(t) = constant is the waveform that results in minimum loss and maximum efficiency.
Solution
Let I be the cycle average value of a periodic i(t). Then i(t) can be expressed as a constant plus a pure alternating component as i(t) = I + i_{ac}(t), where i_{ac}(t) is bipolar and has equal areas under positive halfcycle and negative halfcycle. Let P_{S} be the average power delivered by the source, P_{L} be the average power delivered to the load and P_{R} be the average power dissipated in R. Average powers are equal to corresponding cycle average powers if i(t) lasts for a long time compared to its period. Then,
where T is the period of i_{ac}(t). The cycle average value of a pure alternating waveform with equal magnitude areas under its positive and negative halfcycles will be zero. Therefore, P_{S} = VI W.
Since the instantaneous power in R contains a term proportional to i_{ac}(t), its basic period is T and not 0.5T. This is why the integration in the above step is from 0 to T. The second integral goes to zero since cycle average of a pure alternating component is zero.
The quantity [i_{ac}(t)]^{2} is always positive and hence the second integral in the previous equation will be positivevalued. Therefore,
with the equality sign applicable only when i_{ac}(t) = 0 for all t – i.e., only when i(t) = I, a constant.
Therefore, for a given amount of average current drawn from a DC source, the power loss in the internal resistance and connecting link resistance will be a minimum when the energy is drawn at a constant rate – i.e., power delivered is kept constant by keeping current constant.
The optimum way to draw power from a DC source is by drawing a DC current. Similarly, the optimum way to charge a battery is by delivering a constant current into it. No other current waveform is as energy efficient as the DC current waveform when drawing power from a DC voltage source or delivering power to it.
Example: 6.44
An AC voltage source (i.e., a sinusoidal source) v(t) = 325 sin100πt V delivers 1 kW of average power to a load through a resistance of 5 Ω. This resistance is the sum of the internal resistance of the source and the resistance of cable connecting the load to the source. Assume that the load draws a current i(t) = I_{m} sin(100πt + θ) and that I_{m} and θ can be varied keeping the power delivered by the source at 1 kW itself. Find the amplitude of current, power dissipated in the series resistance and the efficiency of power transfer when (i) θ = –80° (ii) θ = 45° (iii) θ = –30° and (iv) θ = 0°.
Solution
Let v(t) = V_{m} sin ωt and i(t) = I_{m} sin (ωt + θ). Then the average power delivered by the source = (We have derived this many times in earlier sections.). The average power delivered depends on magnitude of phase difference between voltage and current waveforms. But it does not depend on the sign of phase difference.
The power delivered by the source is kept constant at 1000 W in this example. Therefore, I_{m }cosθ = 1000/325 = 3.077 A.
Let P_{R} be the average power dissipated in the resistor. Then,
But I_{m} = 3.077/cosθ. Therefore, P_{R} = 4.734R/cos^{2}θ. Now we evaluate the numbers for the various cases:
 θ = –80°. Then, cosθ = 0.1736, ∴I_{m} = 3.077/0.1736 = 17.72 A, ∴P_{R} = 4.734 × 5/0.1736^{2} = 785.4 W, P_{S} = 1000 W, ∴P_{L} = 214.6 W and efficiency = 21.46 %.
 θ = 45°. Then, cosθ = 0.707, ∴I_{m} = 3.077/0.707 = 4.35 A, ∴P_{R} = 4.734 × 5/0.707^{2} = 47.3 W, P_{S} = 1000 W, ∴P_{L} = 952.7 W and efficiency = 95.27 %.
 θ = –30°. Then, cosθ = 0.866, ∴I_{m} = 3.077/0.1736 = 3.55 A, ∴P_{R} = 4.734 × 5/0.866^{2} = 31.6 W, P_{S} = 1000 W, ∴P_{L} = 968.4 W and efficiency = 96.84 %.
 θ = 0°. Then, cosθ = 1, ∴I_{m} = 3.077 A, ∴P_{R} = 4.734 × 5/1^{2} = 23.7 W, P_{S} = 1000 W, ∴P_{L} = 976.3 W and efficiency = 97.63 %.
Increasing magnitude of phase difference between voltage and current makes power transfer to the load highly inefficient.
Show that the fixed amount of average power delivered by a sinusoidal voltage source through its internal resistance and connecting link resistance to a load will reach the load with minimum loss and maximum efficiency when the current drawn by the load has zero phase difference with respect to the source voltage.
Solution
Let v(t) = V_{m} sin ωt and i(t) = I_{m} sin (ωt + θ). Let the total series resistance in the path be R. Then the average power delivered by the source The average power dissipated in R is 0.5RI_{m}^{2} W (see Example 6.44).
But the power delivered by the source is stated to be fixed. Let this fixed value be P W. Then 0.5 V_{m}I_{m}cosθ = P. Therefore, I_{m} = 2P/(V_{m}cosθ). Therefore, the average power loss in R is
The minimum of this loss takes place when cos^{2}θ is a maximum – i.e., θ = 0° or 180°. θ=180° is relevant when the load is delivering power to source.
Therefore, when a sinusoidal voltage source is delivering power to load, a given amount of source power is transferred to load with minimum losses and maximum efficiency when the load draws current from source at zero phase difference at the source terminal.
This fact has great significance in Electrical Power Distribution systems and Electricity tariff structure.
6.5 EFFECTIVE VALUE (RMS VALUE) OF PERIODIC WAVEFORMS
The cycle average power delivered to the same resistance R by different voltage or current waveforms forms the basis for comparison of their effectiveness in delivering useful power to a load. We can define a measure of effectiveness of a given voltage or current waveform by calculating the cyclic average power that will be delivered to a resistance R and by answering the question – what is the value of a DC voltage or current that will result in same average power in R? The measure defined this way is called the RootMeanSquare Value (rms value) of the waveform. It is also called the Effective Value of the waveform.
Effective value or rms value of a waveform is the value of DC quantity that will produce the same heating effect as that produced by the waveform when it is applied as a voltage across a resistance of 1Ω or as a current through a resistance of 1Ω.
The assumption of 1Ω is a matter of convenience. It could have been R Ω and that will not make any difference since R will get cancelled out when the heating effects are equated.
We can develop an expression for rms value of a periodic waveform as shown in the following. Let x(t) be a periodic waveform with period of T s. We find out the cyclic average power as if x(t) is a voltage signal applied across 1Ω.
Let X_{rms} be the value of DC quantity that will produce same power in 1Ω. Then,
Therefore, finding rms value of a waveform involves squaring the waveform, finding the mean (i.e., average over a cycle) of the squared waveform over a cycle and then finding the root of the mean. Hence the name – root mean square.
Now we can express the average power delivered to a resistance of R Ω by a periodic voltage waveform v(t) applied across it as P = (V_{rms})^{2} / R W. The average power delivered to a resistance of R Ω by a periodic current i(t) flowing through it is P = R(I_{rms})^{2} W.
6.5.1 RMS Value of Sinusoidal Waveforms
Let v(t) = V_{m} sinωt V. We find its rms value as,
Thus, a sinusoidal voltage is only as effective as a DC voltage that is 70.7% of its peak value as far as its heating capability is concerned.
Let v(t) = V_{m} sinωt V and i(t) = I_{m} sin(ωt + θ) A be the voltage across and current through an electrical element as per passive sign convention, respectively. Then we know that the average power delivered to the element is given by 0.5 V_{m}I_{m} cosθW. Now we can express this power equation in terms of rms values of voltage and current.
Two other measures are useful in the context of periodic waveforms. They are cycle average and cycle average of absolute value of the waveform. We denote them by V_{cav} and V_{caav}, respectively.
Note that both in the definitions of rms value in Eqn. 6.51 and average values in Eqn. 6.53, we are free to carry out the integration over any interval of width equal to the period of the waveform. It does not have to be between 0 to T.
A pure alternating waveform will have a cycle average of zero.
The cycle average of absolute value of a pure alternating waveform that has identical positive halfcycle width and negative halfcycle width will be same as the average of positive halfcycle over 0.5T. Hence, the cycle average of absolute value is also called halfcycle average in the case of such a waveform. A sinusoidal waveform is one such waveform. The V_{caav} value for a sinusoidal waveform v(t) = V_{m}sinωt is obtained as follows:
The ratio between the rms value and the halfcycle average value of a periodic waveform with equal halfcycle widths and zero cycle average (i.e., a pure alternating waveform) is defined as its form factor. Therefore, the form factor of a pure sinusoidal waveform
The ratio between peak value and rms value of an alternating waveform with equal halfcycle widths and zero cycle average is defined as its crest factor. The crest factor of a pure sinusoidal waveform is
Example: 6.51
If v(t) is a pure alternating periodic waveform with period T and rms value V_{rms} find the rms value of a new periodic waveform v_{1}(t) = V + v(t) where V is a constant quantity. Thereby, find the rms value of 10+10sin100πt V.
Solution
The component 2Vv(t) represents a pure alternating component with zero cycle average since v(t) is pure alternating waveform and V is a constant. Hence, integration of this term over one cycle period yields zero integral.
Therefore rms value of 10+10sin100πt
Example: 6.52
If v(t) is a periodic waveform with period T, rms value V_{rms} and cycle average value V_{cav}, find the rms value and cycle average value of a new periodic waveform v_{1}(t) = V + v(t), where V is a constant quantity.
v(t) has a nonzero cycle average value. Therefore it can be written as V_{cav} + v_{ac}(t), where v_{ac}(t) is a pure alternating waveform with zero cycle average. Using the result we arrived at in Example 6.51,
∴ V_{ac} _{rms}^{2} = V_{rms}^{2} – V_{cav}^{2}
Now, v_{1}(t) = V + v(t) = V + V_{cav} + v_{ac}(t).
Using the result from Example 6.51 again,
Substituting for V_{ac} _{rms}^{2},
The cycle average of v_{1}(t) = V_{cav} + V
Example: 6.53
Find the rms value of a symmetric square wave of ± 10 V and 50 Hz frequency.
Solution
When a symmetric square wave of amplitude V is squared, the resulting waveform is the same as squaring a DC voltage of V V. Therefore, rms value of a symmetric square wave is same as the amplitude of the wave. Hence, the rms value of square wave voltage in this example is 10 V.
Example: 6.54
Find the rms value, halfcycle value, crest factor and form factor for a symmetric triangular waveform shown in Fig. 6.51.
Fig. 6.51 Symmetric triangle waveform for Example 6.54
Solution
Due to symmetry, the squared waveform needs to be integrated only from 0 to 0.25T. The integral for 0 to T range is four times this value.
Example: 6.55
Find an expression for rms value of the periodic pulse train shown in Fig. 6.52.
Fig. 6.52 Periodic waveform for Example 6.55
Solution
Note the limits of integration.
Example: 6.56
Let v(t) = V_{m} sinωt V. A new voltage waveform is generated from this waveform by a process called halfwave rectification, which is represented mathematically as
Find the rms value and cycle average value of v_{r}(t).
Solution
The area under squared v_{r}(t) over one T will be half the area under squared v(t) since one halfcycle is missing in v_{r}(t). Therefore, rms value of v_{r}(t) will be times the rms value of v(t). Therefore, rms value of v_{r} (t) = 0.5 V_{m} V.
Halfcycle average of v(t) is 2V_{m}/π. Therefore, halfcycle area This area becomes the full cycle area in v_{r}(t) since second half cycle is zerovalued. Therefore, cycle average of v_{r} (t) = V_{m}T/π ÷ T = V_{m}/π V.
6.6 THE POWER SUPERPOSITION PRINCIPLE
Current variables and voltage variables in a linear electric circuit obey the Superposition Theorem. However, instantaneous power in a linear circuit does not follow superposition principle. Consider a circuit driven by two sources v_{S1}(t) and v_{S2}(t) and let the corresponding response components for the voltage across a particular element be v_{11}(t) and v_{12}(t). The corresponding response components of current through that element are i_{11}(t) and i_{12}(t) as per the passive sign convention. Then,
The instantaneous power delivered to the element when v_{S1}(t) is acting alone,
p_{1}(t) = v_{11}(t) i_{11}(t)
The instantaneous power delivered to the element when v_{S2}(t) is acting alone,
p_{2}(t) = v_{12}(t) i_{12}(t)
The instantaneous power delivered to the element when v_{S1}(t) and v_{S2}(t) are acting together,
p(t) should have been equal to p_{1}(t) + p_{2}(t) if instantaneous power obeys superposition principle. Therefore, instantaneous power does not follow superposition principle in a linear circuit. This conclusion is general in nature and there are no exceptions.
However, there are exceptions to this rule when it comes to average power. Let us confine to the case of periodic waveforms that are applied to circuits for sufficient duration such that average power can be taken as cyclic average power. Let the average power delivered to the element in question when v_{S1}(t) is acting alone be P_{1} and the corresponding value when v_{S2}(t) is acting alone be P_{2}. We assume that v_{S1}(t) and v_{S2}(t) have a least common period of T s. This common period need not be the period of any one of them. For instance, let the period of v_{S1}(t) be 1 sec and that of v_{S2}(t) be 1.25 s. Then the value of T will be 5 s since there will be integer number of cycles of both in a 5 s period. Five seconds is the lowest interval that will contain integer number of cycles of both. Cyclic average power can be found by integrating over multiple cycles provided the averaging is done over the same number of cycle periods. Therefore, taking T as the integration interval is permitted in the evaluation of P_{1} and P_{2}.
The average power delivered to the element when both sources are acting together, P, is given by
Hence, in general, average power does not follow superposition principle. However, unlike the case of instantaneous power, the cross power terms in the average power can go to zero under special conditions.
Consider a case where v_{S1}(t) and v_{S2}(t) are two sinusoidal waveforms at different angular frequencies of ω_{1} and ω_{2}. Note that one of them can be zero as a special case indicating a DC or steady input. Assume that ω_{1} < ω_{2}. In this case, T will be an integer multiple of 2π/ω_{1} and another integer multiple of where k_{1} and k_{2} are two integers. It is possible to find a value for T satisfying this condition only if the ratio ω_{1}/ω_{2} is a rational number.
The circuit is linear. Therefore v_{11}(t) and i_{11}(t) will be sinusoids at ω_{1} rad/s and v_{12}(t) and i_{12}(t) will be sinusoids at ω_{2} rad/s. Therefore, both cross power terms – v_{11}(t) i_{12}(t) and v_{12}(t) i_{11}(t) – will involve products of two sinusoidal waveforms of different frequencies – ω_{1} and ω_{2}. By applying trigonometric identities they may be expressed as sinusoidal waveforms of sum frequency and difference frequency. Therefore, v_{11}(t) i_{12}(t) + v_{12}(t) i_{11}(t) will yield two sinusoids – one with a frequency of ω_{1}+ω_{2} rad/s and the second with a frequency of ω_{2} – ω_{1} rad/s.
Both these sinusoidal waveforms will have integer number of cycles within an interval of length T as shown next.
Therefore, the area under these two sinusoidal waveforms over an interval of T s long will be zero. Thus, the cross power terms in the instantaneous power contribute only zero to the average power in this case. Therefore, P = P_{1} + P_{2}. We can extend this analysis to a case involving many sinusoidal sources of different frequencies easily and arrive at the Power Superposition Principle stated in the following:
The average power delivered to an element in a linear circuit excited by sinusoidal sources of different frequencies (including DC, i.e., zero frequency) obeys superposition principle.
6.6.1 RMS Value of a Composite Waveform
We will employ the superposition principle for average power to arrive at an expression for the rms value (i.e., the effective value) of a periodic waveform that is the sum of many sinusoidal periodic waveforms of different frequencies.
Let v(t) = v_{1}(t) + v_{2}(t) +...+ v_{n}(t) be a composite waveform comprising n distinct frequency sinusoidal waveforms. Notice the word distinct used in the last sentence. No two components in the sum can have same frequency value. If there are two sources of same frequency, they have to be combined and treated as a single term.
Now imagine that we apply this signal as a voltage across 1Ω resistor. The average power delivered to the resistor is found by applying power superposition principle as the sum of average power delivered to the same resistor when each source is acting alone. But that will be nothing but the square of rms value of each component waveform. The rms value of v(t) is obtained by finding the square root of average power delivered to the 1Ω resistor. Therefore,
The effective value of a waveform that is the sum of many sinusoidal waveforms with distinct frequencies (including zero frequency) is obtained by taking the square root of sum of squares of rms values of individual components.
Example: 6.61
The voltage applied across an electrical load circuit is v(t) = 10 + 50 sin100πt + 30 sin (150πt – 30°) V and the current delivered to the load circuit is seen to be i(t) = 2.5 + 5 sin (100πt –45°)+ 1.5 sin (150πt – 80°) A. (i) Find the average power delivered to the load circuit, rms value of applied voltage and load current. (ii) What is the period over which averaging was carried out in the last step?
Solution
 All the three components in voltage waveform have distinct frequencies. Similarly, all the three components in current waveform too have distinct frequencies.
Therefore, average power delivered by applying power superposition principle is obtained as
The rms values of voltage and current waveforms are obtained by applying Eqn. 6.62.
ω_{1} = 100π and ω_{2} = 150π

The lowest values possible for k_{1} and k_{2} are 2 and 3 respectively.
Therefore,
Therefore, the period over which averaging was carried out in calculating average power and rms values is 40 ms.
Example: 6.62
Find the rms value of a voltage waveform given by v(t) = 10 sin(100πt – 30°) + 5 cos(100πt + π/6) + 5 cos(150πt – 65°) + 7 sin (300πt – π/3) – 7 cos(300πt – 45°) V.
Solution
The terms in the expression for v(t) contain nondistinct frequencies. Terms with same frequency will have to be combined into a single term first.
Example: 6.63
The current through a 10Ω resistor is seen to be i(t) = 2 + 3 sin 500πt + 1.5 cos 501πt A. (i) Find the amplitude of a symmetric triangle wave current with 50 Hz frequency that will produce same heating in the resistor. (ii) Will the answer arrived at in (i) be correct if the current lasted only for 0.5 s?
 The rms value of a triangular current with a peak value of I_{p} A will be
(see Example 6.53)
This must be equal to the rms value of i(t).
 No. The period over which averaging is done in the calculation of rms value is 500 times the period of sin 500πt. That is, T = 500 × 4ms = 2 s.
Approximating average power by cyclic average power is valid only if the waveform lasts for duration much longer than the period over which averaging is carried out in cyclic average calculation. This is the reason why the frequency information does not figure in the expressions for P and rms values. The effective value of a waveform does not depend on frequency provided the waveform remains applied to the circuit for much longer than its period.
6.7 SUMMARY
 Sinusoidal waveforms are of great importance to electrical and electronic circuits due to the following reasons: (i) They preserve their waveshape in linear circuits. (ii) They render the voltage levels in electrical and electronic systems flexible so that optimisation of subsystem performance in various parts of the system by choosing a suitable voltage level in that part of the system becomes possible. Transformers help us to realise this voltage flexibility. (iii) Periodic nonsinusoidal waveforms as well as a broad class of aperiodic waveforms can be expressed as a sum of sinusoids by Fourier series and Fourier transforms. Hence, circuit solution with such input waveforms can be obtained with relative ease if solution for a sinusoidal input is known.
 A sinusoidal waveform of period T can be expressed as v(t) = A sin (ωt + θ) = A sin (2πt/T + θ) = A sin (2πft + θ), where A is its amplitude, T is its period in s, f = 1/T is its cyclic frequency in s^{1 }(Hertz, Hz) and ω = 2πf/ = 2π/T is its radian frequency or angular frequency in rad/s. The quantity θ in v(t) = A sin (ωt + θ) is defined as the phase of the sinusoidal function.
 The angular difference between similarly located points within a cycle period on two normalised sinusoidal waveforms with same ω is defined as the phase difference between them. The phase difference between two sinusoids is independent of choice of origin in t or ωt axis. The precedence relationship [i.e., which comes after (in a visual sense) which] between them in t or ωt axis too is independent of choice of origin. However, the phase of a sinusoidal waveform depends on the choice of origin in t or ωt axis.
 When a waveform point on a sinusoidal function v_{2}(t) appears after a similarly located point on the waveform of another sinusoidal function v_{1}(t) with same frequency, v_{2}(t) is said to lag v_{1}(t) in phase and θ_{1} – θ_{2} is called a lag phase angle under this condition.
 Similarly, when a waveform point on a sinusoidal function v_{2}(t) appears before a similarly located point on the waveform of another sinusoidal function v_{1}(t) with same frequency, v_{2}(t) is said to lead v_{1}(t) in phase and θ_{2} – θ_{1} is called a lead phase angle under this condition.
 Phase lag is not necessarily a phase delay and phase lead is not necessarily a phase advance. Phase lag does not necessarily imply time delay and phase lead does not necessarily imply time advance.
 Instantaneous power delivered to a twoterminal element, p(t) = v(t) i(t), where v(t) and i(t) are the element variables defined as per passive sign convention. The sum of p(t) delivered to all elements in an isolated circuit will be zero.
 p(t) delivered to an element is a nonconstant function of time in general. If v(t) and i(t) are periodic waveforms with period T and zero average value over a cycle period, p(t) will be a periodic waveform with period 0.5T and may have a nonzero average value over its cycle period of 0.5T.
 The energy delivered to the element will also be a function of time. The value of total energy delivered to the element at endofcycle points will fall on a straight line with a slope equal to the average value of p(t) waveform over its cycle period.
 The Cycle Average Power in the context of periodic waveforms is defined as the average of p(t) over one cycle period and is denoted by P. Average power, P_{av} over a time interval (t_{2}–t_{1}) is defined as Average Power, P_{av} = Cycle Average Power, P, if the waveforms remain applied to the circuit for sufficient duration compared to their period.
 Effective value or rms value of a waveform is the value of DC quantity that will produce the same heating effect as that produced by the waveform when it is applied as a voltage across a resistance of 1Ω or as a current through a resistance of 1Ω. If x(t) is a periodic waveform with period T, its rms value is given by
 Let v(t) = V_{m} sinωt V and i(t) = I_{m} sin(ωt + θ) A be the voltage across and current through an electrical element as per passive sign convention. Then rms value of v(t) is V_{m}/√2, rms value of current is I_{m}/√2 and the average power delivered to the element is V_{rms}I_{rms} cosθ W.
 The average power delivered to an element in a linear circuit excited by sinusoidal sources of different frequencies (including DC, i.e., zero frequency) obeys superposition principle.
 Let v(t) = v_{1}(t) + v_{2}(t) + ... + v_{n}(t) be a composite waveform comprising n distinct frequency sinusoidal waveforms. Then,
6.8 PROBLEMS
 A periodic waveform is the sum of three sinusoidal waveforms of period T, 0.5T and 0.2T respectively. Will this periodic waveform preserve its waveshape when differentiated and integrated?
 What is the amplitude of v(t) = 5 sinωt – 3 cosωt and what is its phase in degrees with respect to v_{1}(t) = 4 sin(ωt – π/5)?
 Is the composite waveform v(t) = 2sinωt + 3cos√2ωt periodic? If yes, what is its period?
 Is the composite waveform v(t) = 2 sin200πt + 4 cos(200.0001πt – 45°) periodic? If yes, what is its period and cyclic frequency?
 The voltage across a linear load is v(t) = 100 sin(100πt – 25°) V. If the load current i(t) is found to lag v(t) by 36° and i(0) is 2 A find i(t) as a function of time.
 What is the phase of the first derivative of a sinusoidal waveform with respect to the original waveform?
 What is the phase of the indefinite integral of a sinusoidal waveform with respect to the original waveform?
 If v_{1}(t) = V_{1} sin ωt and v_{2}(t) = V_{2} sin (ωt + θ), what is the value of θ such that (i) v_{1}(t) +v_{2}(t) has a minimum amplitude, (ii) v_{1}(t) +v_{2}(t) has a maximum amplitude and (iii) v_{1}(t) +v_{2}(t) has an amplitude of (V_{1} + V_{2})/2.
 If v_{1}(t) = V_{1} sin ωt, v_{2}(t) = V_{2} sin (ωt + θ) and v_{3}(t) = v_{1}(t) +v_{2}(t) = V_{3} sin (ωt + θ), find θ such that (V_{1, }V_{2}, V_{3}) is a Pythagorean triplet and find ɸ under this condition in terms of V_{1} and V_{2}.
 Derive an expression for phase angle of v_{2}(t) with respect to v_{1}(t) if v_{1}(t) = a sin ωt + b cos ωt and v_{2}(t) = c sin ωt + d cos ωt.
 Find the amplitude, frequency, angular frequency and phase of the following sinusoidal functions of time.
 Find the phase of x(t) and y(t) with respect to a sinωt function, phase difference between x(t) and y(t) and the lead/lag relationship between them for the following cases:
 If x(t) = 2sin50πt + 3 cos55πt + 3 sin60πt, find the period and cyclic frequency of x(t).
 A linear circuit contains 6 twoterminal elements. The sum of instantaneous power in five of them at a particular instant is found to be 17 W. What is the nature of the remaining element and what is the instantaneous power delivered to it at that instant?
 A linear element has v(t) = 10 sin10πt V and i(t) = 3 cos (10πt – 45°) A as per passive sign convention. (i) Is the element a resistor? If not, why? (ii) Can the element be a simple element like an inductor or a capacitor or must it be a composite element? (iii) Find the average rate of growth of energy dissipated in the element. (iv) Is this average rate accurate enough if the element remained powered up only for 0.43 s? If not, calculate the average rate at which energy was dissipated in it.
 The square wave in Fig. 6.81 is the voltage across an element and the triangle wave is the current through it. (i) Find and plot the instantaneous power delivered to the element and the energy delivered to it as functions of time for t > 0. (ii) Calculate the cycle average power.
Fig. 6.81
 Calculate the average power an element with v(t) same as the square wave in Fig. 6.81 and i(t) = (5+the triangle wave in Fig. 6.81) for t≥0.
 Show that average power in an isolated circuit is a conserved quantity.
 Show that cyclic average power in an isolated circuit is a conserved quantity.
 An active circuit delivers i(t) = 10 sin300πt A into the positive terminal of a 12 V battery. The battery has an internal resistance of 0.2 Ω. (i) What is the average power delivered to the 12 V source inside the battery? (ii) What is the average power loss in the internal resistance of the battery?
 The voltage across a linear load is v(t) = 325 sin(100πt – 25°) V. If the load current i(t) is found to lead v(t) by 36° and the average power delivered to the load is 250 W, find i(t) as a function of time.
 Derive an expression for average power if v(t) = a sin ωt + b cos ωt and i(t) = c sin ωt + d cos ωt in terms of a, b, c and d.
 A sinusoidal voltage source of 230 V rms delivers 1 kW of average power to a load. What is the minimum possible amplitude of load current?
 The periodic ramp voltage waveform in Fig. 6.82 is applied across a resistor of 10 Ω. Find the average power dissipated in it.
Fig. 6.82
 The absolute value of v(t) shown in Fig. 6.83 is applied across a 5 Ω resistor. Find the average power delivered to it.
Fig. 6.83
 A pure sinusoidal voltage v(t) = 100 sin100πt V is applied across a 20 Ω resistor. (i) Find the amount of energy that is moving back and forth between the voltage source and the resistor. (ii) Find the time required such that this amount is less than 1 % of the energy dissipated in the resistor.
 A linear load draws i(t) = 5 sin200πt A from a voltage source v(t) = 200 cos 200πt V. (i) Find the average power delivered to the load. (ii) Find the amount of energy that is moving back and forth between the voltage source and the load.
 A linear composite load draws i(t) = 5 sin(100πt – 30°) A from a voltage source v(t) = 200 sin100πt V. (i) Find the average power delivered to the load. (ii) Find the amount of energy that is moving back and forth between the voltage source and the load. (iii) Find the time required such that this amount is less than 1% of the energy dissipated in the load.
 A 48V battery with a series resistance of 0.1 Ω delivers the current shown in Fig. 6.84 to a power electronic load through a connecting wire resistance of 0.05 Ω. (i) Find the cycle average and rms value of i(t). (ii) Find the average power delivered by the source and received by the load. (iii) What is the waveshape and value of i(t) that would have resulted in minimum loss in the system with source delivering same average power?
Fig. 6.84
 What is the ratio between rms values of a symmetric square wave and a sinusoidal waveform with same amplitude?
 What are the ratios between rms values and halfcycle average values of a symmetric triangle waveform and a sinusoidal waveform with same amplitude?
 The rms value of a periodic waveform v(t) with a frequency of 1kHz is found to be 10 V. What is the rms value of v_{1}(t) which has same amplitude and waveshape as that of v(t), but has a frequency of 10 Hz?
 The rms value of V_{m} sin ωt is V_{m}/√2for any value of ω. Explain why ω value does not affect the rms value.
 If the average power delivered by a 50 Hz sinusoidal voltage to a resistor of 10Ω is 2 W, what is the average power delivered to the same resistor by another sinusoidal voltage source with same amplitude and 5 kHz frequency?
 Show that if V_{rms} is the rms value of a periodic waveform v(t), the rms value of av(t) is aV_{rms} where a is a real constant.
 A sinusoidal voltage source v(t) = 100 sin20πt is applied across a 10Ω resistor for 0.42 s from t = 0. (i) Find the cyclic average power. (ii) Find the average power delivered during the connection duration. (iii) Why are the two values different?
 A nonsinusoidal periodic voltage waveform with rms value of 100 V is applied to a 10Ω resistor. What is the average power delivered to the resistor?
 If v(t) = V_{m}sinωt + 0.5 V_{m} cos3ωt V and i(t) = I_{m} sin(ωt – 45°) – 2I_{m}cos(3ωt – 45°) A are the terminal variables of a load, find the average power delivered to the load.
 (a) If v_{1}(t) = 5 sin 100πt and v_{2}(t) = 5 sin 100πt, find the rms value of v_{1}(t) + v_{2}(t), average power that will be delivered to a 10Ω resistor by v_{1}(t) + v_{2}(t), and the energy that will be delivered to 10Ω resistor in 0.5 s. (b) If v_{1}(t) = 5 sin 100πt and v_{2}(t) = 5 sin 100.000001πt, repeat part (a). Can the energy delivered in 0.5 s be found from average power in this case? Why are the rms values so different in (a) and (b) for such a small change in frequency of one component?
 A current i(t) = 10 sin 200πt flows through a resistor of 5 Ω from t = 0 to t = 32 ms. Explain why energy delivered to the resistor is not equal to average power multiplied by duration of connection. Calculate the actual value of energy dissipated in the resistor.
 A sinusoidal voltage source v(t) = V_{m} sinωt V delivers a fixed amount of average power P through its internal resistance of R to a load. The load current has a general form of
Show that the fixed amount of average power delivered by the source reaches the load with maximum efficiency and minimum loss, when i(t) is a pure sinusoidal waveform at ω rad/s with zero phase difference from v(t), i.e., all a’s and b’s are zerovalued except a_{1}
 Show that the rms value of a periodic waveform x(t) and its absolute value x(t) are equal.
 The voltage appearing across a power electronic load and the current drawn by it are shown in Fig. 6.85. (i) Find the average power delivered to the load. (ii) Find the cycle average value and rms value of v(t) and i(t).
Fig. 6.85
 Find the value of δ such that the periodic waveform shown in Fig. 6.86 will have the same rms value as that of a sinusoidal waveform with same peak voltage of V_{p}.
Fig. 6.86
 Find the cycle average and rms value of the periodic waveform in Fig. 6.87.
Fig. 6.87
 If v(t) across an element is 10 sin100πt V and i(t) in that element is 2 + 3sin200πt + 2 cos300πt A, find the average power delivered?
 If v(t) across an element is 5 + 10 sin100πt V and i(t) in that element is 2 + 3sin100πt + 2 cos100πt +2 sin300πt A, find the average power delivered?
 Calculate the form factor and crest factor for the waveform shown in Fig. 6.88.
Fig. 6.88
 A rectifiertype voltmeter reads the rms value of a sine wave by measuring the halfcycle average of the waveform and graduating the meter scale after accounting for the form factor of the waveform. What will be the meter reading if (i) a 10 V peak sinusoidal waveform is applied to it, (ii) a 10 V peak symmetrical square waveform is applied to it and (iii) a 10 V peak symmetric triangular waveform is applied to it?
 Find the form factor and crest factor of the current waveform shown in Fig. 6.89.
Fig. 6.89
 The halfcycle average of a periodic voltage waveform is 25 V. Form factor is 1.2. Find the power delivered to a 10 Ω resistor when this waveform is applied across it?
 The cycle average of a periodic voltage waveform is 25 V. Its rms value is 35 V. Find the power delivered to a 5 Ω resistor when this waveform is applied to it after removing its DC component?
 Find the rms value of v(t) = 20 + 20 sin100πt + 35 cos (100πt – 35°) + 23 sin (250πt – 0.2π) + 21 cos (250πt + 45°) V.
 If the current in an element as per passive sign convention is i(t) = 3 + 2 sin(100πt – 36°) + 3.5 cos (100πt – 71°) + 1.15 sin (250πt – 0.45π) + 1.05 cos (250πt) A when the voltage waveform in Problem 24 is applied across it, find the average power delivered to it.
 One period of two periodic voltage waveforms v_{1}(t) and v_{2}(t) is shown in Fig. 6.810. Find the power delivered by (v_{1}(t) + v_{2}(t)) to a 10 Ω resistor. Explain why average power does not satisfy superposition principle in this case.
Fig. 6.810
 v_{1}(t) is a composite periodic waveform containing many sinusoidal waveforms of distinct frequencies. v_{2}(t) is another composite periodic waveform containing many sinusoidal waveforms of distinct frequencies. v_{1}(t) and v_{2}(t) do not share a sinusoidal waveform component of same frequency. Show that the rms values of [v_{1}(t) + v_{2}(t) ] and [v_{1}(t) – v_{2}(t) ] will be the same.