Masters G.M. Renewable and Efficient Electric Power Systems

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48 BASIC ELECTRIC AND MAGNETIC CIRCUITS

annual energy saved. This is the cost of conserved energy (CCE) and is

described more carefully in Section 5.4. CCE is deﬁned as follows

CCE =

annual cost of saved electricity($/yr)

annual electricity saved (kWhr/yr)

P · CRF(i, n)

kWhr/yr

where P is the extra cost of the conservation feature (heavier duty

wire in this case), and CRF is the capital recovery factor (which means

your annual loan payment on $1 borrowed for n years at interest rate i.

What would be the “cost of conserved energy” CCE (cents/kWhr) if

the building (and wiring) is being paid for with a 7-%, 20-yr loan with

CRF = 0.0944/yr. How does that compare with the cost of electricity

that you don’t have to purchase from the utility at 10¢ /kWhr?

1.11. Suppose a photovoltaic (PV) module consists of 40 individual cells wired

in series, (a). In some circumstances, when all cells are exposed to the sun

it can be modeled as a series combination of forty 0.5-V ideal batteries, (b).

The resulting graph of current versus voltage would be a straight, vertical

20-V line as shown in (c).

(a) 40-cell PV module

0.5 V

40 cells

(b) 40 cells in sun

20 V

Figure P1.11

a. When an individual cell is shaded, it looks like a 5- resistor instead

of a 0.5-V battery, as shown in (d). Draw the I-V curve for the PV

module with one cell shaded.

(d) one cell shaded

0.5 V

5 Ω

0.5 V

39 cells

1 cell

shaded

0.5 V

38 cells

2 cells

shaded

(e) two cells shaded

5 Ω

Figure P1.11

PROBLEMS 49

b. With two cells shaded, as in (e), draw the I-V curve for the PV module

on the same axes as you have drawn the full-sun and 1-cell shaded

I-V lines.

1.12. If the photovoltaic (PV) module in Problem 1.11 is connected to a 5-

load, ﬁnd the current, voltage, and power that will be delivered to the load

under the following conditions:

PV module

5 Ω

Load

−

Figure P1.12

a. Every cell in the PV module is in the sun.

b. One cell is shaded.

c. Two cells are shaded.

Use the fact that the same current and voltage ﬂows through both the PV

module and the load so solve for I, V, and P.

1.13. When circuits involve a source and a load, the same current ﬂows through

each one and the same voltage appears across both. A graphical solution

can therefore be obtained by simply plotting the current-voltage (I-V) rela-

tionship for the source onto the same axes that the I-V relationship for the

load is plotted, and then ﬁnding the crossover point where both are sat-

isﬁed simultaneously. This is an especially powerful technique when the

relationships are nonlinear.

For the photovoltaic module supplying power to the 5-W resistive load

in Problem 1.12, solve for the resulting current and voltage using the

graphical approach:

a. For all cells in the sun.

b. For one cell shaded

c. For two cells shaded

CHAPTER 2

FUNDAMENTALS OF ELECTRIC POWER

2.1 EFFECTIVE VALUES OF VOLTAGE AND CURRENT

When voltages are nice, steady, dc, it is intuitively obvious what is meant when

someone says, for example, “this is a 9-V battery.” But what does it mean to say

the voltage at the wall outlet is 120-V ac? Since it is ac, the voltage is constantly

changing, so just what is it that the “120-V” refers to?

First, let us describe a simple sinusoidal current:

i = I

cos(ωt + θ) (2.1)

where i is the current, a function of time; I

is the magnitude, or amplitude, of the

current; ω is the angular frequency (radians/s); and θ is the phase angle (radians).

Notice that conventional notation uses lowercase letters for time-varying voltages

or currents (e.g., i and v), while capitals are used for quantities that are constants

(or parameters), (e.g., I

or V

rms

). Also note that we just as easily could have

described the current with a sine function instead of cosine. A plot of (2.1) is

showninFig.2.1.

The frequency ω in (2.1) is expressed in radians per second. Equally common

is to express the frequency f in hertz (Hz), which are “cycles per second.” Since

there are 2π radians per cycle, we can write

ω = 2π

(

radians/cycle

)

· f

(

cycles/s

)

= 2πf (2.2)

Renewable and Efﬁcient Electric Power Systems. By Gilbert M. Masters

ISBN 0-471-28060-7

 2004 John W iley & Sons, Inc.

52 FUNDAMENTALS OF ELECTRIC POWER

Magnitude,

amplitude

= 2p

Figure 2.1 Illustrating the nomenclature for a sinusoidal function.

The sinusoidal function is periodic—that is, it repeats itself—so we can also

describe it using its period, T :

T = 1/f ( 2.3)

Thus, the sinusoidal current can have the following equivalent representations:

i = I

cos(ωt + θ) = I

cos(2πf t + θ) = I

cos



2π

t +θ



(2.4)

Suppose we have a portion of a circuit, consisting of a current i passing

through a resistance R asshowninFig.2.2.

The instantaneous power dissipated by the resistor is

p = i

R(2.5)

In (2.5), power is given a lowercase symbol to indicate that it is a time-varying

quantity. The average value of power dissipated in the resistance is given by

avg

= (i

)

avg

R = I

eff

R(2.6)

In (2.6) an effective value of current, I

eff

, has been introduced. The advantage

of deﬁning the effective value this way is that the resulting equation for average

Figure 2.2 A time-varying current i through a resistance R.

EFFECTIVE VALUES OF VOLTAGE AND CURRENT 53

power dissipated looks very similar to the instantaneous power described by (2.5).

This leads to a deﬁnition of the effective value of current given below:

eff



)

avg

= I

rms

(2.7)

The effective value of current is the square root of the mean value of cur-

rent squared. That is, it is the root-mean-squared, or rms, value of current. The

deﬁnition given in (2.7) applies to any current function, be it sinusoidal or oth-

erwise.

To ﬁnd the rms value of a function, we can always work it out formally using

the following:

rms



)

avg





(t)dt (2.8)

Oftentimes, however, it is easier simply to graph the square of the function and

determine the average by inspection, as the following example illustrates.

Example 2.1 RMS value of a Square Wave. Find the rms value of a square-

wave current that jumps back and forth from 0 to 2 A as shown below:

Solution We need to ﬁnd the square root of the average value of the square of

the current. The waveform for current squared is

The average value of current squared is 2 by inspection (half the time it is zero,

half the time it is 4):

rms



)

avg

√

54 FUNDAMENTALS OF ELECTRIC POWER

= cos

−1

0.5

Figure 2.3 The average value of the square of a sinusoid is 1/2.

Let us derive the rms value for a sinusoid by using the simple graphical

procedure. If we start with a sinusoidal voltage

v = V

cos ωt (2.9)

The rms value of voltage is

rms



)

avg



cos

ωt)

avg

= V



(cos

ωt)

avg

(2.10)

Since we need to ﬁnd the average value of the square of a sine wave, let us graph

y = cos

ωt as has been done in Fig. 2.3.

By inspection of Fig. 2.3, the mean value of cos

ωt is 1/2. Therefore, using

(2.10), the rms value of a sinusoidal voltage is

rms

= V



√

(2.11)

This is a very important result:

The rms value of a sinusoid is the amplitude divided by the square root of 2.

Notice this conclusion applies only to sinusoids! When an ac current or voltage

is described (e.g., 120 V, 10 A), the values speciﬁed are always the rms values.

Example 2.2 Wall outlet voltage. Find an equation like (2.1) for the 120-V,

60-Hz voltage delivered to your home.

IDEALIZED COMPONENTS SUBJECTED TO SINUSOIDAL VOLTAGES 55

Solution From (2.11), the amplitude (magnitude, peak value) of the voltage is

√

rms

= 120

√

2 = 169.7V

The angular frequency ω is

ω = 2πf = 2π60 = 377 rad/s

The waveform is thus

v = 169.7 cos 377t

It is conventional practice to treat the incoming voltage as having zero phase

angle, so that all currents will have phase angles measured relative to that refer-

ence voltage.

2.2 IDEALIZED COMPONENTS SUBJECTED TO SINUSOIDAL

VOLTAGES

2.2.1 Ideal Resistors

Consider the response of an ideal resistor to excitation by a sinusoidal voltage

as shown in Fig. 2.4.

The voltage across the resistance is the same as the voltage supplied by

the source:

v = V

cos ωt =

√

rms

cos ωt =

√

2V cos ωt (2.12)

Notice the three ways that the voltage has been described: using the amplitude

of the voltage V

, the rms value of voltage V

rms

, and the symbol V which, in

this context, means the rms value. We will consistently use current I or voltage

V (capital letters, without subscripts) to mean the rms values of that current

or voltage.

The current that will pass through a resistor with the above voltage imposed

will be

i =

cos ωt =

√

rms

cos ωt =

√

cos ωt (2.13)

=√2

cos w

−

Figure 2.4 A sinusoidal voltage imposed on an ideal resistance.

56 FUNDAMENTALS OF ELECTRIC POWER

Since the phase angle of the resulting current is the same as the phase angle of

the voltage (zero), they are said to be in phase with each other. The rms value

of current is therefore

rms

= I =

√

2V/R

√

(2.14)

Notice how simple the result is: The rms current I is equal to the rms voltage V

divided by the resistance R. We have, in other words, a very simple ac version

of Ohm’s law:

V = RI (2.15)

where V and I are rms quantities.

Now let’s look at the average power dissipated in the resistor.

avg

= (vi)

avg

= [

√

2 V cos ωt ·

√

2 I cos ωt]

avg

= 2VI(cos

ωt)

avg

(2.16)

The average value of cos

ωt is 1/2. Therefore,

avg

= 2VI ·

= VI (2.17)

In a similar way, it is easy to show the expected alternative formulas for average

power are also true:

avg

= VI = I

R =

(2.18)

Notice how the ac problem has been greatly simpliﬁed by using rms values

of current and voltage. You should also note that the power given in (2.18) is

the average power and not some kind of rms value. Since ac power is always

interpreted to be average power, the subscript in P

avg

is not usually needed.

Example 2.3 ac Power for a Lightbulb. Suppose that a conventional incan-

descent lightbulb uses 60 W of power when it supplied with a voltage of 120 V.

Modeling the bulb as a simple resistance, ﬁnd that resistance as well as the

current that ﬂows. How much power would be dissipated if the voltage drops

to 110 V?

Solution Using (2.18), we have

R =

(120)

= 240 

IDEALIZED COMPONENTS SUBJECTED TO SINUSOIDAL VOLTAGES 57

and

I =

120

= 0.5A

When the voltage sags to 110 V, the power dissipated will be

P =

(110)

240

= 50.4W

2.2.2 Idealized Capacitors

Recall the deﬁning equation for a capacitor, which says that current is propor-

tional to the rate of change of voltage across the capacitor. Suppose we apply an

ac voltage of V volts (rms) across a capacitor, as shown in Fig. 2.5.

The resulting current through the capacitor will be

i = C

= C

(

√

2V cos ωt) =−ωC

√

2V sin ωt (2.19)

If we apply the trigonometric identity that sin x =−cos(x + π/2),weget

i =

√

2ωCV cos



ωt +



(2.20)

There are several things to note about (2.20). For one, the current waveform

is a sinusoid of the same frequency as the voltage waveform. Also note that there

is a 90

◦

phase shift (π/2 radians) between the voltage and current. The current is

said to be leading the voltage by 90

◦

. That the current leads the voltage should

make some intuitive sense since charge must be delivered to the capacitor before

it shows a voltage. The graph in Fig. 2.6 also suggests the idea that the current

peaks 90

◦

before the voltage peaks.

Finally, writing (2.20) in terms of (2.1) gives

i =

√

2ωCV cos



ωt +



= I

cos(ωt + θ) =

√

2I cos(ωt + θ) (2.21)

=√2

cos w

−

Figure 2.5 An ac voltage V , applied across a capacitor.