Notes for course EE1.1 Circuit Analysis 2004-05

TOPIC 11 – POWER IN AC CIRCUITS

Objectives:

• To consider averaging a signal

• To determine the average power in a load

• To consider power relationships in a resonant circuit

• To define and work with RMS values of voltage and current

1 AVERAGES

1.1 Average of a general function

Consider a waveform xT such as the one shown that has finite duration:

We denote the average value of a waveform xT(t) by: <xT(t)>

The precise definition of average (or mean) value is as follows: the average of xT(t) over theinterval [0, T] is that constant <xT(t)> that encloses exactly the same area as xT(t)

Because <xT(t)> is a constant, its enclosed area is that of a rectangle with height <xT(t)> and lengthT:

Mathematically, area is simply the definite integral, so we can equate the two areas:

Thus, we can solve for <xT(t)>:

�

xT t( ) = 1T

xT t( )dt0

T

∫

If xT(t) is formed by combining simple shapes, such as rectangles, triangles or trapezia, then it is notnecessary to carry out the integration; it is probably simpler to interpret the integral as the areaunder the xT(t) curve

The process of "taking the average" is an operator; it accepts a waveform and produces a constant:the average value

It can be shown that this operator is linear, ie:

for any two waveforms x(t) and y(t) and any pair of numbers a and b

We also note that the average of a constant is the constant itself, that is,

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We can perform an important simplification for the average value of a periodic waveform

In this case, the average over a very long time interval is the same as the average over one periodbecause the waveform repeats itself; hence we can integrate just over one period from t = 0 to T0:

Hence, we can find the area enclosed by one period of the waveform and divide by the period tofind the average value

1.2 Average of sinusoids and related waveforms

Consider the waveform of the function cos(ωt + θ):

The average value over one period can be seen to be zero because the areas above and below thehorizontal axis having equal magnitude and their signs are positive and negative, respectively

Thus, the net area over one full period is zero

Because sin(ωt + θ) = cos(ωt + θ – 90°), we see that its average is also zero; in fact, the average ofany sinusoid-with any frequency and with any phase-is zero

Another waveform whose average we will encounter is the square of the general sinusoid, oneperiod of which is shown:

We have drawn it with a phase angle of zero for convenience of representation

We would expect the average value to be 0.5 because the waveform is symmetric with respect to ahorizontal line at that value

Clearly the waveform is nonnegative because it is a squared function

To calculate the averages of such functions, we use the following trigonometric identity:

For a general time-varying sinusoid with phase shift, we have

We can now apply the property of linearity and our knowledge of the average of a constant and of ageneral sinusoid as follows:

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A similar method may be used to show that:

<sin2(ωt + θ)> = 1/2:

We summarise these results in a table:

2 AVERAGE POWER IN A LOAD

2.1 Basic expression

Now we can determine the average power absorbed by a general 2-terminal load for the case ofsinusoidal voltages and currents:

We assume that the load consists of any combination of the circuit elements we have studied thusfar: resistors, inductors, capacitors, dependent sources, and/or independent sources

We assume, however, that each of the independent sources (if there are any) is sinusoidal at thefrequency ω

We allow the voltage to have, the general phase angle θ

If the entire circuit in which the load is connected, including the load itself, is stable, then aftersome time has passed the current will be a sinusoidal waveform – the circuit will be operating in theAC steady state

Then the current will have some phase angle, which we have shown as θ + φ; thus, φ is zero whenthe phase of the current is exactly the same as that of the voltage

If φ > 0, we say the current leads the voltage; if φ < 0, we say the current lags the voltage

We show one complete period of both v(t) and i(t) for the case in which φ = –90°

The crest, or peak, of the current waveform comes after – or lags – the crest of the voltagewaveform

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The instantaneous power absorbed by the load in the above set-up is given by:

The average power is the average value of the instantaneous power:

We can apply the trigonometric identity:

wirh

x = ωt + θ + φ and y = ωt + θ

Hence, we obtain a very important result in AC power theory:

The constant cos(φ) is called the power factor; it is usually written P F; the adjective leading orlagging is added to indicate the sign of φ

Example 18

Find the power absorbed by the following circuit for values of the voltage source phase angle α =0°, 90°, and –90°.

Solution

We first convert to phasor form:

We now find

�

IAn application of KVL and Ohm's law gives

For α = 0°,

�

I = 0

For α = 90°:

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For α = –90°:

For α = 0°, Im = 0 A, so P = 0 W

For α = 90°, Im = 3 A and φ = –90°, P = (1/2)VmImcos(φ) = (1/2) × 12 V × 3 A × cos(–900) = 0 W

For α = –90°, Im = 3 A also, and φ = 0°; thus, P = (1/2) × 12 V × 3 A × cos(0°) = 18 W.

Notice that for α = 90°, the average power is zero even though neither the current nor the voltage iszero

2.2 Average Power Absorbed by the Passive Elements

Now let's look at the power consumed by our three basic 2-terminal elements: the resistor, thecapacitor, and the inductor:

We assume that:

�

V =Vm∠θ

I = Im∠θ + φ

For the resistor:

�

Z jω( ) = R

I = VZ jω( )

= Vm∠θR

= VmR

∠θ

Im = VmR

φ = 0

P = 12VmIm cosφ = Vm

2

2RNote that the power factor cos(φ) is unity

For the inductor:

�

Z jω( ) = jωL = ωL∠90

I = VZ jω( )

= Vm∠θωL∠90

= VmωL

∠θ − 90

Im = VmωL

φ = −90

P = 12VmIm cosφ = 0

For the capacitor:

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�

Z jω( ) = 1jωC

= 1ωC

∠− 90

I = VZ jω( )

= ωCVm∠90

Im = ωCVm φ = 90

P = 12VmIm cosφ = 0

Unlike the resistor, the ideal inductor and the ideal capacitor do not dissipate any power; they aresometimes referred to as lossless elements

2.3 Power relationships in a resonant circuit

Let's consider a tuned-circuit in the time domain:

We assume that v(t) = Vmcos(ωt) V and look at the AC steady-state currents indicated and thepower absorbed by each of the elements:

We use phasor techniques to solve for the forced response

Resistor current is IR = (Vm∠0)/R = Vm/R, so the time-domain steady-state response is:

Inductor current is IL = Vm/jωL = (Vm/ωL) ∠–90°, so the time-domain response is

Capacitor current is IC = jωCVm = ωCVm∠+90°, so

These three currents for one period of positive time are as follows:

The resistor current is exactly in phase (φ = 0°) with the applied voltage

The phase of the inductor current is –90° with respect to the voltage

The phase of the capacitor current is +90° with respect to the voltage

It follows that the inductor and capacitor currents have a relative phase shift of 1800

This implies that the capacitor supplies some of the current for the inductor during one half-period,and that the opposite is true during the next half-period

Let's now look at the instantaneous steady-state power absorbed by each element:

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For the resistor, we have:

There is a constant part and a time-varying part

Computing the average, we find that the average of the time-varying part is zero; thus,

A graph of PR(t) is shown for one period:

Notice that the resistor power is always positive – the resistor only absorbs power

Also, notice that the period of the power waveform is one-half the period of the voltage (andcurrent).

The AC steady-state inductor power is:

The AC steady-state capacitor power is:

Because the inductor and capacitor currents are so similar, let's look at the associated powerstogether on the same plot:

Power flows from the capacitor to the inductor during one half-period and conversely during thenext

The two maximum values are different, in general, with the difference being supplied by the voltagesource

But let's look at the special case in which the two maxima are equal; in this case, we set (1/2)ωCVm2

= Vm2/(2ωL)

If we think of keeping the inductance and the capacitance values fixed and varying the frequency,this will occur when

a condition called parallel resonance

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At resonance, the peak energy stored by the capacitor is equal and opposite to that stored by theinductor

The inductor and capacitor swap this energy back and forth, and the voltage source must supplyonly the energy absorbed by the resistor

The average value of energy absorbed by both the inductor and by the capacitor is zero

3 RMS VALUES

We now consider how effective AC voltages and currents are in delivering power to loads

Consider a possible experiment:

On the left, we have a general time-varying voltage source v(t) connected to a resistive load

On the right, we show a DC voltage source connected to the same load; the arrow indicates that thevoltage is adjustable

The DC voltage is adjusted so that, for some value which we will call Vrms, the average powerabsorbed by the resistor is exactly the same as that absorbed in the left hand figure

We can equate the average power in the resistor in the two circuits:

Solving for Vrms we have:

We call this quantity the root-mean-square value of v(t) because we must first square v(t), thencompute the average (or mean) and then take the square root

Note that the rms value is a constant, whereas v(t) is a time-varying waveform

The rms value is a scalar measure of how effective the waveform is in delivering average power toa load; hence rms values are often referred to as effective values

We similarly show:

In the special case in which the voltage v(t) is a sinusoid, we can write v(t) = Vmcos(ωt + φ)

We can quickly compute

�

v2 t( ) =Vm2 2 , so

Similarly,

We note that these results are only for sinusoids

We conclude this section by deriving the RMS values for sime typical waveforms:

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Squaring the waveform gives exactly the same waveform because 12 = 1 and 02 = 0

Thus, the average of the squared waveform is 1/2

Taking the square root, we obtain Xrms = 1/√2

Because (–1)2 = 1, the square of this waveform is unity, which has the average value of 1

Taking the square root gives Xrms = 1

The nonzero portion of this waveform is the sine function:

Hence, Xrms = 1/2

The period of this waveform is one-half of the previous one; other than that, the computation is thesame

Thus

�

x2 t( ) =1 2 and Xrms = 1/√2

In the expression x{t) = Xmcos(ωt + θ), the value Xm is referred to as the peak value

The rms and peak values of a sinusoid are realted by:

�

Vrms = Vm2

An AC voltage of 230 Vrms will develop the same power in a resistor as a DC voltage of 230 Vapplied to the same load

The peak value of the AC voltage in this case is 230 × √2 = 325.3 V

4 POWER IN RMS TERMS

The equation we derived for average power in a load with sinusoidal voltage and current can be re-written using rms voltage and current values:

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Thus, we can drop the factor of 1/2 if we use rms voltage and current units

Previously, the sinusoidal voltage v(t) = Vmcos(ωt + θ), for example, has the phasor V = Vm∠θ

In some situations, phasors are scaled by the factor of 1/√2 and thus write V = Vrms∠θ where Vrms =

Vm/√2

Hence, power is given by the product of the magnitudes of the voltage and current phasors time thepower factor (no factor of 1/2 is needed)

Ohm's law,

�

V = Z jω( )I is unaffected by using rms voltage and current since both voltage and

current peak values are divided by √2

Example 19

For the 2-terminal load shown, find the rms values of the terminal voltage, the resistor voltage, thecapacitor voltage, and the terminal current; also compute the power factor and the average powerabsorbed by the load:

Solution

The peak voltage of 157 V, when divided by √2, gives an rms value of 110 V

The radian frequency of 337 rad/s corresponds to 377/(2!) = 60 Hz

The phasor equivalent (using rms phasors for the voltages and currents) is as follows:

The rms value of the current is:

�

I = 110∠0

3− j4= 110∠0

5∠− 53.1= 22∠53.1 Arms

The resistor voltage is, therefore:

�

VR = 3× 22∠53.1 = 66∠53.1 Vrms

The capacitor voltage is obtained as the product:

�

VC = − j4( ) × 22∠53.1 = 4∠− 90 × 22∠53.1 = 88∠− 36.9 Vrms

The angle of the current is 53.1°, positive, so:

�

Power Factor = cos 53.1( ) = 0.4 leading

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The average power absorbed by the load is:

�

P =VrmsIrms × PF = 110 × 22 × 0.4 = 968 W

5 CONCLUSIONS

In this topic, we have considered the important topic of power in more detail than hitherto. Webegan by considering the process of taking the average of a signal. We then determined expressionsfor the average power in a load and considered power relationships in a resonant circuit. Finally,we defined RMS values of voltage and current and used them to derive simpler power relationships.