ANSWERS TO FOCUS ON CONCEPTS QUESTIONS ... TO FOCUS ON CONCEPTS QUESTIONS 1. (d) According to 2 R s / (Equation 20.15c), the average power is proportional to the square of the rms

CHAPTER 23 ALTERNATING CURRENT

CIRCUITS

ANSWERS TO FOCUS ON CONCEPTS QUESTIONS

1. (d) According to 2rms

/P V R (Equation 20.15c), the average power is proportional to the

square of the rms voltage. Tripling the voltage causes the power to increase by a factor of

32 = 9.

2. Irms = 1.9 A

3. (b) The current Irms through a capacitor depends inversely on the capacitive reactance XC,

as expressed by the relation Irms = Vrms/ XC (Equation 23.1). The capacitive reactance

becomes infinitely large as the frequency goes to zero (see Equation 23.2), so the current

goes to zero.

4. (e) According to C1/ 2X f C (Equation 23.2) and

L2X f L (Equation 23.4),

doubling the frequency f causes XC to decrease by a factor of 2 and XL to increase by a

factor of 2.

5. Irms = 1.3 A

6. (a) The component of the phasor along the vertical axis is V0 sin 2 f t (see the drawing that

accompanies this problem), which is the instantaneous value of the voltage.

7. (b) The instantaneous value of the voltage is the component of the phasor that lies along the

vertical axis (see Sections 23.1 and 23.2). This vertical component is greatest in B and least

in A, so the ranking is (largest to smallest) B, C, A.

8. (d) In a resistor the voltage and current are in phase. This means that the two phasors are

colinear.

9. (c) Power is dissipated by the resistor, as discussed in Section 20.5. On the other hand, the

average power dissipated by a capacitor is zero (see Section 23.1).

10. Irms = 2.00 A

11. (a) When the rms voltage across the inductor is greater than that across the capacitor, the

voltage across the RCL combination leads the current (see Section 23.3).

1178 ALTERNATING CURRENT CIRCUITS

12. (d) Since Irms= Vrms/Z (Equation 23.6), the current is a maximum when the impedance Z is a

minimum. The impedance is 22

L CZ R X X (Equation 23.7), and it has a

minimum value when XC = XL = 50 .

13. (c) The inductor has a very small reactance at low frequencies and behaves as if it were

replaced by a wire with no resistance. Therefore, the circuit behaves as two resistors, R1 and

R2, connected in parallel. The inductor has a very large reactance at high frequencies and

behaves as if it were cut out of the circuit, leaving a gap in the connecting wires. The circuit

behaves as a single resistance R2 connected across the generator. The situation at low

frequency gives rise to the largest possible current, because the effective resistance of the

parallel combination is smaller than the resistance R2.

14. (a) The capacitor has a very small reactance at high frequencies and behaves as if it were

replaced by a wire with no resistance. Therefore, the circuit behaves as two resistors, R1 and

R2, connected in parallel. The capacitor has a very large reactance at low frequencies and

behaves as if it were cut out of the circuit, leaving a gap in the connecting wires. Therefore,

the circuit behaves as a single resistor R1 connected across the generator. The situation at

high frequencies gives rise to the largest possible current, because the effective resistance of

the parallel combination is smaller than the resistance R1.

15. (e) At low frequencies, the capacitor has a very large reactance. In the series circuit, this

large reactance gives rise to a large impedance and, hence, a small current. The parallel

circuit has the larger current, because current can flow through the inductor, which has a

small reactance at low frequencies.

16. (b) The resonant frequency f0 is given by 0

1

2f

LC (Equation 23.10). It depends only

on C and L, and not on R.

17. f0 = 1.3 103 Hz

18. (d) The resonant frequency f0 is given by 0

1

2f

LC (Equation 23.10). When a second

capacitor is added in parallel, the equivalent capacitance increases (see Section 20.12).

Therefore, the resonant frequency decreases.

Chapter 23 Problems 1179

CHAPTER 23 ALTERNATING

CURRENT CIRCUITS

PROBLEMS

1. SSM REASONING As the frequency f of the generator increases, the capacitive

reactance XC of the capacitor decreases, according to C

1

2X

f C (Equation 23.2), where

C is the capacitance of the capacitor. The decreasing capacitive reactance leads to an

increasing rms current Irms, as we see from rmsrms

C

VI

X (Equation 23.1), where Vrms is the

constant rms voltage across the capacitor. We know that Vrms is constant, because it is equal

to the constant rms generator voltage. The fuse is connected in series with the capacitor, so

both have the same current Irms. We will use Equations 23.1 and 23.2 to determine the

frequency f at which the rms current is 15.0 A.

SOLUTION Solving Equation 23.2 for f, we obtain

C

1

2f

C X (1)

In terms of the rms current and voltage, Equation 23.1 gives the capacitive reactance as

rmsC

rms

VX

I . Substituting this relation into Equation (1) yields

rms

6C rmsrms

rms

1 1 15.0 A9470 Hz

2 2 2 63.0 10 F 4.00 V2

If

C X CVVC

I

2. REASONING When two capacitors of capacitance C are connected in parallel, their

equivalent capacitance CP is given by P

2C C C C (Equation 20.18). We will find the

capacitive reactance XC of the equivalent capacitance from CP

1

2X

f C (Equation 23.2).

Because the capacitors are the only devices connected to the generator, the rms voltage Vrms

across the capacitors is equal to the rms voltage of the generator. Therefore, the capacitive

reactance XC is related to the rms voltage of the generator and the rms current Irms in the

circuit by rms rms C

V I X (Equation 23.1).


SOLUTION Substituting CP = 2C into C

P

1

2X

f C (Equation 23.2) and solving for C,

we obtain

C

P C

1 1 1 1 or

2 2 2 4 4X C

f C f C fC fX (1)

Solving rms rms C

V I X (Equation 23.1) for XC yields rmsC

rms

VX

I . Substituting this result

into Equation (1), we obtain

7rms

rmsC rms

rms

1 1 0.16 A 8.7 10 F

4 4 4 610 Hz 24 V4

IC

VfX fVf

I

3. REASONING The reactance XC of a capacitor is given as C

1

2X

f C (Equation 23.2),

where f is the frequency of the ac current and C is the capacitance of the capacitor. We note

that XC is inversely proportional to f for a given value of C. Therefore, we will be able to

solve this problem by applying Equation 23.2 twice, once for each value of the frequency

and each time with the same value of the capacitance.

SOLUTION Applying Equation 23.2 for each value of the frequency, we obtain

C, 870 C, 460870 460

1 1 and

2 2X X

f C f C

Dividing the equation on the left by the equation on the right and noting that the unknown

capacitance C is eliminated algebraically, we find that

C, 870 870 460

C, 460 870

460

1

2

1

2

X f C f

X f

f C

Solving for the reactance at a frequency of 870 Hz gives

460C, 870 C, 460

870

460 Hz68 36

870 Hz

fX X

f

4. REASONING The rms voltage Vrms and current Irms in a capacitor are related according to

rms rms CV I X (Equation 23.1). XC is the capacitive reactance C

1

2X

f C

(Equation 23.2), where f is the frequency in hertz (Hz) and C is the capacitance of the

capacitor. We will apply these equations to the capacitor when connected to each of the


generators. Although we are not given a value for the capacitance, we will see that it is

eliminated algebraically from the calculation.

SOLUTION Substituting Equation 23.2 for the capacitive reactance into Equation 23.1, we

obtain

rms rms C rms

1

2V I X I

f C

(1)

Applying Equation (1) to the capacitor when connected to generator 1 and generator 2, we

have

rms rms rms rms1 1 2 21 2

1 1 and

2 2V I V I

f C f C

Dividing the equation on the right by the equation on the left, we note that the capacitance C

is eliminated algebraically and find that

rms rms 2 rms 1 rms 12 2 2 2

rms rms 1 2 rms 2 rms1 1 1 1

/ 2 2

/ 2 2

V I f C I f C I f

V I f C f C I f I

(2)

Solving Equation (2) for the voltage of the second generator gives

3 3rms rms 11 2

rms 3 322 rms 1

2.0 V 85 10 A 3.4 10 Hz3.3 V

5.0 10 Hz 35 10 A

V I fV

f I

5. SSM REASONING The rms current in a capacitor is Irms

= Vrms

/XC, according to

Equation 23.1. The capacitive reactance is XC = 1/(2 f C), according to Equation 23.2. For

the first capacitor, we use C = C1 in these expressions. For the two capacitors in parallel, we

use C = CP, where C

P is the equivalent capacitance from Equation 20.18 (C

P = C

1 + C

2).

Taking the difference between the currents and using the given data, we can obtain the

desired value for C2. The capacitance C

1 is unknown, but it will be eliminated algebraically

from the calculation.

SOLUTION Using Equations 23.1 and 23.2, we find that the current in a capacitor is

IV

X

V

f CV f C

rms

rms

C

rms

rms

1 22

/ b g

Applying this result to the first capacitor and the parallel combination of the two capacitors,

we obtain

I V f C I V f C C1 rms

Single capacitor

Combination rms

Parallel combination

and 2 21 1 2

c h

Subtracting I

1 from I

Combination, reveals that


I I V f C C V f C V f CCombination 1 rms rms rms

2 2 21 2 1 2

c h Solving for C

2 gives

CI I

V f2

6

2

0 18

2 4402 7 10

Combination 1

rms

A

24 V Hz F

..b g b g

6. REASONING The rms current in the circuit is given by Irms

= Vrms

/XC (Equation 23.1),

where Vrms

is the rms voltage of the generator, and XC = 1/(2 f C) is the capacitive

reactance (see Equation 23.2). For a given voltage, smaller reactances lead to greater

currents.

When two capacitors are connected in parallel, the equivalent capacitance CP is given by

CP = C

1 + C

2 (Equation 20.18), where C

1 and C

2 are the individual capacitances. Therefore,

CP is greater than either C

1 or C

2. Thus, when the capacitors are connected in parallel, the

greater capacitance leads to a smaller reactance (C is in the denominator in Equation 23.2),

which in turn leads to a greater current. As a result, the current delivered by the generator

increases when the second capacitor is connected in parallel with the first capacitor.

The capacitance of a parallel plate capacitor is given by C = 0A/d (Equation 19.10), where

is the dielectric constant of the material between the plates, 0 is the permittivity of free

space, A is the area of each plate, and d is the separation between the plates. When the

capacitor is empty, = 1, so that C = Cempty

. Thus, the capacitance increases when the

dielectric material is inserted.

SOLUTION Using Equation 23.1 to express the current as Irms

= Vrms

/XC and Equation 23.2

to express the capacitive reactance as XC = 1/(2 f C), we have for the current that

IV

X

V

f CV f C

rms

rms

C

rms

rms

1 22

/ b g

Applying this result to the case where the empty capacitor C

1 is connected alone to the

generator and to the case where the “full” capacitor C2 (which contains the dielectric

material) is connected in parallel with C1, we obtain

1 1 2

1, rms rms 1 P, rms rms P

alone and in parallel

2 and 2

C C C

I V f C I V f C

Dividing the two expressions gives

I

I

V f C

V f C

C

C

P PP, rms

1, rms

rms

rms

2

21 1


According to Equation 20.18, the equivalent capacitance of the two capacitors in parallel is

CP = C

1 + C

2, so that the result for the current ratio becomes

I

I

C C

C

C

C

P, rms

1, rms

1 2

1

2

1

1

Since the capacitance of a filled capacitor is given by Equation 19.10 as C =

0A/d, we find

that

I

I

A d

A d

P, rms

1, rms

1 10

0

/

/

Solving for IP, rms

gives

I IP, rms 1, rms

A A 1 0 22 1 4 2 1 1b gb gb g. . .

7. REASONING The capacitance C is related to the capacitive reactance XC and the

frequency f via Equation 23.2 as C = 1/(2fXC). The capacitive reactance, in turn is related

to the rms-voltage Vrms

and the rms-current Irms

by XC = V

rms/I

rms (see Equation 23.1). Thus,

the capacitance can be written as C = Irms

/(2fVrms

). The magnitude of the maximum charge

q on one plate of the capacitor is, from Equation 19.8, the product of the capacitance C and

the peak voltage V.

SOLUTION

a. Recall that the rms-voltage Vrms

is related to the peak voltage V by Vrms

= V

2. The

capacitance is, then,

CI

f V F

HGIKJ rms

rms2

A

2 HzV

2

F

3 0

750140

6 4 10 6.

.

b g

b. The maximum charge on one plate of the capacitor is

q CV 6 4 10 140 9 0 106 4. .F V Cc hb g

8. REASONING AND SOLUTION Equations 23.1 and 23.2 indicate that the rms current in a

capacitor is I V X /C

, where V is the rms voltage and X f CC1 2/ b g. Therefore, the

current is I V f C 2 . For a single capacitor C C1, and we have

I V f C 2

1

For two capacitors in series, Equation 20.19 indicates that the equivalent capacitance can be


obtained from C C C–1 –1 –1 1 2

, which can be solved to show that C C C C C 1 2 1 2

/c h.

The total series current is, then,

I V f C V fC C

C Cseries

FHG

IKJ2 2 1 2

1 2

The series current is one-third of the current I. It follows, therefore, that

I

I

V fC C

C C

V f C

C

C C

C

C

series or

FHG

IKJ

2

2

1

32

1 2

1 2

1

2

1 2

1

2

For two capacitors in parallel, Equation 20.18 indicates that the equivalent capacitance is

C C C 1 2

. The total current in this case is

I V f C V f C Cparallel

2 21 2

c h The ratio of I

parallel to the current I in the single capacitor is

I

I

V f C C

V f C

C C

C

C

C

parallel 3

2

2

21 1

1

2

1 2

1

1 2

1

2

1

c h

9. SSM REASONING The rms current can be calculated from Equation 23.3,

I V Xrms rms L

/ , provided that the inductive reactance is obtained first. Then the peak value

of the current I0 supplied by the generator can be calculated from the rms current I

rms by

using Equation 20.12, I I0

2rms

.

SOLUTION At the frequency of f 620 Hz , we find, using Equations 23.4 and 23.3, that

X f LL

Hz)(8.2 2 2 620 10 ( –3 H) = 32

IV

Xrms

rms

L

10.0 V

32 A

0 31.

Therefore, from Equation 20.12, we find that the peak value I

0 of the current supplied by the

generator must be

I I0 2 2 0.31 A 0.44 A

rms b g

10. REASONING The rms voltage Vrms and current Irms in an inductor are related according to

rms rms LV I X (Equation 23.3). XL is the capacitive reactance L 2X f L

(Equation 23.4), where f is the frequency in hertz (Hz) and L is the inductance of the

inductor. Since we have values for Vrms, f, and L, we can use these equations to calculate

the unknown current Irms.


SOLUTION Substituting Equation 23.4 for the inductive reactance into Equation 23.3, we

obtain

rms rms L rms2V I X I f L (1)

Solving Equation (1) for Irms, we find that

rmsrms rms rms

55 V2 or 0.17 A

2 2 650 Hz 0.080 H

VV I f L I

f L

11. REASONING The rms voltage Vrms across the inductor is given by

rms rms LV I X (Equation 23.3), where Irms is the rms current in the circuit, and XL is the

inductive reactance. The inductor is the only circuit element connected to the generator, so

the rms voltage across the inductor is equal to the rms generator voltage: Vrms = 15.0 V.

SOLUTION Solving Equation 23.3 for XL, we obtain

rmsL

rms

15.0 V24.6

0.610 A

VX

I

12. REASONING The inductance L of the inductor determines its inductive reactance XL

according to L

2X fL (Equation 23.4), where f is the frequency of the generator. When

the inductor is connected to the terminals of the generator, the rms voltage Vrms of the

generator drives an rms current Irms that depends upon the inductive reactance via

rmsL

rms

VX

I (Equation 23.3). We note that the generator frequency is given in kHz, where

1 kHz = 103 Hz, and the rms current is given in mA, where 1 mA = 10−3 A.

SOLUTION Solving L

2X fL (Equation 23.4) for L yields

L

2

XL

f (2)

Substituting rmsL

rms

VX

I (Equation 23.3) into Equation (1), we find that

rms

rms rmsL

3 3rms

39 V0.020 H

2 2 2 2 7.5 10 Hz 42 10 A

V

I VXL

f f fI


13. REASONING Since the capacitor and the inductor are connected in parallel, the voltage

across each of these elements is the same or V VL C . Using Equations 23.3 and 23.1,

respectively, this becomes I X I Xrms L rms C

. Since the currents in the inductor and

capacitor are equal, this relation simplifies to X XL C . Therefore, we can find the value

of the inductance by equating the expressions (Equations 23.4 and 23.2) for the inductive

reactance and the capacitive reactance, and solving for L.

SOLUTION Since X XL C , we have

21

2

f L

f C

Therefore, the value of the inductance is

Lf C

1

4

1

4 60.0 100 176

2 2 2 ( ( H = 176 mH

Hz) 40.0 F)2 –6.

14. REASONING The current in L1 is given by Equation 23.3 as I

rms = V

rms/X

L, where

XL = 2 f L

1 (Equation 23.4) is the inductive reactance of L

1. This current does not depend in

any way on L2 and exists whether or not L

2 is present.

The current delivered to the parallel combination is the sum of the currents delivered to each

inductance and is, therefore, greater than either individual current. The current in L2 is given

by Irms

= Vrms

/XL (Equation 23.3), where X

L = 2 f L

2 is the inductive reactance of L

2

according to Equation 23.4. This current does not depend in any way on L1 and exists

whether or not L1 is present.

SOLUTION Using Equation 23.3 to express the current as Irms

= Vrms

/XL and Equation 23.4

to express the inductive reactance as XL = 2 f L, we have for the current that

IV

X

V

f Lrms

rms

L

rms 2

Applying this result to the case where L

1 or L

2 is connected alone to the generator, we obtain

IV

X

V

f LI

V

X

V

f L

L L

1, rms

rms

L

rms

alone

2, rms

rms

L

rms

alone1 2

and 2 2

1 2

The current delivered to L

1 alone is

IV

f L1, rms

rms V

Hz H A

2

240

2 2200 6 0 102 9

13 b gc h.

.


The current delivered to the parallel combination of L1 and L

2 is the sum of that delivered

individually to each inductor and is

I I IV

f L

V

f L

V

f L LP, rms 1, rms 2, rms

rms rms rms

V

Hz H H A

FHG

IKJ

FHG

IKJ

2 2 2

1 1

240

2 2200

1

6 0 10

1

9 0 104 8

1 2 1 2

3 3

b g . ..

15. SSM REASONING

a. The inductive reactance XL depends on the frequency f of the current and the inductance

L through the relation XL = 2 f L ( Equation 23.4). This equation can be used directly to

find the frequency of the current.

b. The capacitive reactance XC depends on the frequency f of the current and the capacitance

C through the relation XC = 1/(2 f C ) ( Equation 23.2). By setting XC = XL as specified in

the problem statement, the capacitance can be found.

c. Since the inductive reactance is directly proportional to the frequency, tripling the

frequency triples the inductive reactance.

d. The capacitive reactance is inversely proportional to the frequency, so tripling the

frequency reduces the capacitive reactance by a factor of one-third.

SOLUTION a. The frequency of the current is

34L

3

2.10 101.11 10 Hz

2 2 30.0 10 H

Xf

L

(23.4)

b. The capacitance is

9

4 3C

1 16.83 10 F

2 2 1.11 10 Hz 2.10 10C

f X

(23.2)

c. Since XL = 2 f L , tripling the frequency f causes XL to also triple:

3 3L

= 3 2.10 10 = 6.30 10X

d. Since XC = 1/(2 f C ) , tripling the frequency f causes XC to decrease by a factor of 3:

3 21C 3

= 2.10 10 = 7.00 10X


16. REASONING AND SOLUTION Equations 23.3 and 23.4 indicate that the rms current in a

single inductance L1 is I

1 = V/X

L1, where V is the rms voltage and X

L1 = 2fL

1. Therefore,

the current is I V f L1 1

2 / c h. Similarly, the current in the second inductor connected

across the terminals of the generator is I V f L2 2

2 / c h. The total current delivered by

the generator is the sum of these two values:

I I IV

f L

V

f Ltotal

1 2

1 22 2

But this same total current is delivered to the single inductance L, so it also follows that

I V f Ltotal

/ 2b g. Equating the two expressions for Itotal

shows that

V

f L

V

f L

V

f L L L L2 2 2

1 1 1

1 2 1 2

or

Using this result, we determine the value of L as follows:

1 1 1 0 030 0 060

0 030 0 0601 2

1 2

1 2L L L

LL L

L L

or

H H

H H0 020 H

. .

. .

b gb g.

17. SSM REASONING The voltage supplied by the generator can be found from Equation

23.6, V I Zrms rms

. The value of Irms

is given in the problem statement, so we must obtain

the impedance of the circuit.

SOLUTION The impedance of the circuit is, according to Equation 23.7,

Z R X X 2 2 2275 3 60 10( – ) .L C

2 2( ) (648 – 415 )

The rms voltage of the generator is

V I Zrms rms

A ) = 83.9 V ( . )( .0 233 3 60 102

18. REASONING As discussed in Section 23.3, the power factor is

22

L C

cosR

R X X

, where R is the resistance, XL is the inductive reactance, and XC

is the capacitive reactance of the circuit. The inductive reactance is given by L 2X f L

(Equation 23.4), where f is the frequency in hertz (Hz) and L is the inductance. The

capacitive reactance is given by C

1

2X

f C (Equation 23.2), where C is the capacitance.


SOLUTION Using Equation 23.4 and Equation 23.2 to substitute for the reactances, we

find that the power factor is

2 222L C

2

2 3

6

cos

12

2

47.0

147.0 2 2550 Hz 4.00 10 H

2 2550 Hz 2.00 10 F

0.819

R R

R X XR f L

f C

19. SSM REASONING We can use the equations for a series RCL circuit to solve this

problem provided that we set C 0 X since there is no capacitor in the circuit. The

current in the circuit can be found from Equation 23.6, rms rms

V I Z , once the impedance of

the circuit has been obtained. Equation 23.8, L C

tan ( ) /X X R , can then be used (with

C0X ) to find the phase angle between the current and the voltage.

SOLUTION The inductive reactance is (Equation 23.4)

L2 2 106 Hz 0.200 H 133X f L

The impedance of the circuit is

2 2 2 2 2 2L C L

( – ) (215 ) (133 ) 253 Z R X X R X

a. The current through each circuit element is, using Equation 23.6,

rmsrms

234 V0.925 A

Z 253

VI

b. The phase angle between the current and the voltage is, according to Equation 23.8 (with

C0X ),

–1L C L 133 tan = 0.619 or tan (0.619) 31.8

215

X X X

R R


20. REASONING The phase angle is given by tan = (XL – X

C)/R (Equation 23.8). When a

series circuit contains only a resistor and a capacitor, the inductive reactance XL is zero, and

the phase angle is negative, signifying that the current leads the voltage of the generator.

The impedance is given by Equation 23.7 with XL = 0 , or Z R X 2

C

2 . Since the

phase angle and the impedance Z are given, we can use these relations to find the

resistance R and XC.

SOLUTION Since the phase angle is negative, we can conclude that only a resistor and a

capacitor are present. Using Equations 23.8, then, we have

tan tan . .

X

RX R RC

C or 75 0 3 73b g

According to Equation 23.7, the impedance is

Z R X 192 2C

2

Substituting XC = 3.73R into this expression for Z gives

192 3 73 192

192

14 949 7

2 2 2

2

R R

R

.

..

b g b g

b g

= 14.9 R or 14.9 R2 2

Using the fact that X

C = 3.73 R, we obtain

X

C = 3.73 (49.7 ) = 185

21. REASONING For a series RCL circuit the total impedance Z and the phase angle are

given by

22 L C

L C (23.7) tan (23.8)

X XZ R X X

R

where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance.

In the present case, there is no capacitance, so that C

0 X . Therefore, these equations

simplify to the following:

2 2 LL

(1) tan (2)X

Z R XR

We are given neither R nor XL. However, we do know the current and voltage when only

the resistor is connected and can determine R from these values using rms

rms

VR

I

(Equation 20.14). In addition, we know the current and voltage when only the inductor is


connected and can determine XL from these values using rmsL

rms

VX

I (Equation 23.3). Since

the generator frequency is fixed, this value for XL also applies for the series combination of

the resistor and the inductor.

SOLUTION With only the resistor connected, Equation 20.14 indicates that the resistance

is

rms

rms

112 V224

0.500 A

VR

I

With only the inductor connected, Equation 23.3 indicates that the inductive reactance is

2rmsL

rms

112 V2.80 10

0.400 A

VX

I

a. Using these values for R and XL in Equations (1) and (2), we find that the impedance is

222 2 2

L 224 2.80 10 359 Z R X

b. The phase angle between the current and the voltage of the generator is

21 1L L 2.80 10

tan or tan tan 51.3224

X X

R R

22. REASONING Since, on the average, only the resistor consumes power, the average power

dissipated in the circuit is 2rms

P I R (Equation 20.15b), where Irms is the rms current and R

is the resistance. The current is related to the voltage V of the generator and the impedance Z

of the circuit by rms rms

/I V Z (Equation 23.6). Thus, the average power can be written as

2 2/P V R Z . The impedance of the circuit is (see Equation 23.7) 22

L CZ R X X

2 2C

R X , since there is no inductor in the circuit. Therefore, the expression for the

average power becomes 2 2

2 2 2C

V R V RP

Z R X

SOLUTION The capacitive reactance is

C 6

1 12400

2 2 60.0 Hz 1.1 10 FX

f C

(23.2)

The average power dissipated is


22

2 2 2 2C

120 V 2700 3.0 W

2700 2400

V RP

R X

23. REASONING From Figure 23.11 we see that V V V V0

2 2

L C R

2c h . Since VL = 0 V

(L = 0 H), and we know V0 and V

R, we can use this equation to find V

C.

SOLUTION Solving the equation above for VC gives

V V VC R

2 V V V 0

2 2 245 24 38b g b g

24. REASONING The rms current Irms in the circuit in part c of the drawing is equal to the

rms voltage Vrms divided by the impedance Z or

rmsrms

VI

Z (Equation 23.6). The impedance

Z of a series RC circuit is given by Equation 23.7 with XL = 0 , since there is no

inductance in the circuit; 2 2C

Z R X The resistance R is known and the capacitive

reactance XC can be obtained from the relation XC = 1/(2 f C) (Equation 23.2), where C is

the capacitance and f is the frequency. The capacitance, however, is related to the time

constant of the circuit in part a of the drawing. The time constant of an RC circuit is the

time for the capacitor to lose 63.2% of its initial charge (see the discussion in Section

20.13), and it is equal to the product of the resistance R and the capacitance C; RC

(Equation 20.21).

SOLUTION Substituting 2 2C

Z R X into rmsrms

VI

Z gives

rms rmsrms 2 2

C

V VI

Z R X

(1)

Substituting XC = 1/(2 f C) (Equation 23.2) into Equation (1) allows us to write the rms

current in the circuit in part c of the drawing as follows:

Substituting C = R (Equation 20.21) into Equation (1), we arrive at an expression for the

rms current:

rms rmsrms 2 2 2

2C 1

2

V VI

R XR

f C


25. SSM REASONING We can use the equations for a series RCL circuit to solve this

problem, provided that we set XL 0 since there is no inductance in the circuit. Thus,

according to Equations 23.6 and 23.7, the current in the circuit is I V R Xrms rms C

/ 2 2 .

When the frequency f is very large, the capacitive reactance is zero, or XC 0 , in which

case the current becomes I f V Rrms rms

large( ) / . When the current Irms

in the circuit is

one-half the value of I frms

large )( that exists when the frequency is very large, we have

rms

rms

1

(large ) 2

I

I f

We can use these expressions to write the ratio above in terms of the resistance and the

capacitive reactance. Once the capacitive reactance is known, the frequency can be

determined.

SOLUTION The ratio of the currents is

I

I f

V R X

V R

R

R X

R

R X

Crms

rms

rms

rms C Clarge

or( )

/

/

2 2

2 2

2

2 2

1

2

1

4

Taking the reciprocal of this result gives

R X

R

X

R

2 2

2

2

24 1 4

C Cor

Therefore,

X

R

C 3

According to Equation 23.2, X f CC1 2/ b g, so it follows that

X

R

f C

R

C 1 2

3/ b g

Thus, we find that

rms rmsrms

2 22 2

22

4

1

2 2

24 V0.78 A

18 18

2 380 Hz 3.0 10 s

V VI

RR R

f C f


fRC

1

2 3

1

2 85 3 b gc h4.0 10 F270 Hz

–6

26. REASONING For a series RCL circuit the total impedance Z and the phase angle are

given by

22 L C

L C (23.7) tan (23.8)

X XZ R X X

R

where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance.

In the present case, there is no inductance, so L

0 X . Therefore, these equations

simplify to the following:

2 2 CC

(1) tan (2)X

Z R XR

Values are given for Z and , so we can solve for R and XC.

SOLUTION Using the value given for the phase angle in Equation (2), we find that

CC

tan tan 9.80 0.173 or 0.173 (3)X

X RR

Substituting this result for XC into Equation (1) gives

2 22 2 2

C0.173 1 0.173Z R X R R R

Solving for R reveals that

2

2 2

4.50 10 443

1 0.173 1 0.173

ZR

Using this value for R in Equation (3), we find that

C0.173 0.173 443 76.6 X R

27. REASONING The instantaneous value of the generator voltage is V(t) = V0 sin 2 ft, where

V0 is the peak voltage and f is the frequency. We will see that the inductive reactance is

greater than the capacitive reactance, XL > X

C, so that the current in the circuit lags the

voltage by /2 radians, or 90. Thus, the current in the circuit obeys the relation

I(t) = I0 sin (2 ft 2), where I

0 is the peak current.

SOLUTION

a. The instantaneous value of the voltage at a time of 1.20 10–4

s is


3 40

( ) sin 2 32.0 V sin 2 1.50 10 Hz 1.20 10 s 29.0 VV t V f t

Note: When evaluating the sine function in the expression above, be sure to set your

calculator to the radian mode.

b. The inductive and capacitive reactances are

3 3L

2 2 1.50 10 Hz 7.20 10 H 67.9X f L (23.4)

C 3 6

1 116.1

2 2 1.50 10 Hz 6.60 10 FX

f C

(23.2)

Since XL is greater than X

C, the current lags the voltage by /2 radians. Thus, the

instantaneous current in the circuit is I(t) = I0 sin (2 ft /2), where I

0 = V

0/Z. The

impedance Z of the circuit is

2 2 22

L C0 67.9 16.1 51.8Z R X X (23.7)

The instantaneous current is

0 12

3 4 12

sin 2

32.0 Vsin 2 1.50 10 Hz 1.20 10 s 0.263 A

51.8

VI f t

Z

28. REASONING The rms voltages across the inductor L and the capacitor C are given,

respectively, by L, rms rms L

V I X (Equation 23.3) and C, rms rms C

V I X (Equation 23.1),

where Irms is the rms current in the circuit, L

2X f L (Equation 23.4) is the inductive

reactance of the inductor, and C

1

2X

f C (Equation 23.2) is the capacitive reactance of

the capacitor. We know the frequency f of the generator, but we are not given the rms

current in the circuit. We will make use of Equations 23.3 and 23.1 to eliminate the

unknown current Irms, and then solve for the rms voltage across the inductor.

SOLUTION Solving Equation 23.1 for Irms gives C, rms

rmsC

VI

X . Substituting this relation

into Equation 23.3, we obtain

C, rms L

L, rms rms LC

V XV I X

X (1)


Substituting Equations 23.2 and 23.4 for the reactances into Equation (1) yields

2C, rms L C, rms

L, rms C, rmsC

22

1

2

V X V f LV V f LC

X

f C

Therefore, when the rms voltage across the capacitor is VC, rms = 2.20 V, the rms voltage

across the inductor is

2

L, rms C, rms

2 3 6L, rms

2

2.20 V 2 375 Hz 84.0 10 H 5.80 10 F 5.95 V

V V f LC

V

29. REASONING A resistance R and an inductance L are connected in series to the generator,

but there is no capacitance in the circuit. Therefore, the impedance Z of the circuit is given

by 22

L CZ R X X (Equation 23.7), where XL is the inductive reactance of the

inductor, and the capacitive reactance XC is zero:

22 2 2

L L0 or R X Z R X Z (1)

The impedance Z is related to the rms current Irms and the rms generator voltage Vrms by

rms rmsV I Z (Equation 23.6). The rms voltage VL, rms across the inductor is given by

L, rms

rmsL

VI

X (Equation 23.3). We will determine the inductive reactance from the

generator frequency f and the inductance by means of L

2X f L (Equation 23.4).

SOLUTION Solving rms rms

V I Z (Equation 23.6) for Z, we obtain

rms

rms

VZ

I (2)

Substituting Equation (2) into Equation (1) yields

2 2 rmsL

rms

VR X

I (3)

Substituting L, rms

rmsL

VI

X (Equation 23.3) into Equation (3), we obtain

2 2 2 2rms rms LL L

L, rms L, rms

L

= or V V X

R X R XV V

X

(4)


Squaring both sides of Equation (4) and solving for R2, we find that

2 22

2 2 2 2 2 2rms L rms L rms2

L, rms L, rms L, rms

or or 1L L L

V X V X VR X R X R X

V V V

(5)

Taking the square root of both sides of Equation (5) yields

2

rmsL 2

L, rms

1V

R XV

(6)

Substituting

L2X f L (Equation 23.4) into Equation (6), we obtain the resistance R:

22rms2 2

L, rms

8.0 V2 1 2 130 Hz 0.032 H 1 76

2.6 V

VR f L

V

30. REASONING The inductance L and the capacitance C of a series RCL circuit determine

the resonant frequency f0 according to

0

1

2f

LC (23.10)

As we see from Equation 23.10, the smaller the inductance L, the larger the resonant

frequency, and the larger the inductance, the smaller the resonant frequency. Therefore, in

part (a) we will use the largest frequency to determine the minimum inductance, and in part

(b) we will use the smallest frequency to find the maximum inductance. We note that

1 MHz = 1×106 Hz.

SOLUTION a. Squaring both sides of Equation 23.10 and solving for L, we obtain

20 2 2 2

0

1 1 or

4 4f L

LC f C (1)

The minimum inductance is obtained when the resonant frequency is greatest, so Equation

(1) gives

5

22 6 11

11.7 10 H

4 9.0 10 Hz 1.8 10 F

L

b. Using Equation (1) once more, this time with the smallest resonant frequency, yields the

maximum inductance:

5

22 6 11

18.8 10 H

4 4.0 10 Hz 1.8 10 F

L


31. SSM REASONING The resonant frequency f0 of a series RCL circuit depends on the

inductance L and capacitance C through the relation 0

1

2f

LC (Equation 23.10). Since

all the variables are known except L, we can use this relation to find the inductance.

SOLUTION Solving Equation 23.10 for the inductance gives

5

2 2 22 3 9

0

1 12.7 10 H

4 4 690 10 Hz 2.0 10 F

Lf C

32. REASONING The average power P dissipated in the circuit is rms rms

cosP I V

(Equation 23.9), where Vrms is rms voltage of the generator, Irms is the rms current in the

circuit, and cos ϕ is the power factor. The angle ϕ is the angle between the current and

voltage phasors and can be determined from L Ctan /X X R (Equation 23.8), where

XL is the inductive reactance, XC is the capacitive reactance, and R is the resistance of the

circuit. In Equation 23.9, we have a value for the average power P , the current Irms and we

can obtain a value for the angle ϕ from Equation 23.8 and the fact that the circuit is at

resonance. Therefore, Equation 23.9 can be used to determine the voltage Vrms.

SOLUTION Solving Equation 23.9 for the voltage, we have

rms rms rmsrms

cos or cos

PP I V V

I

(1)

At resonance in a series RCL circuit we know that L CX X . Therefore, Equation 23.8

becomes

1L C 0.00tan 0.00 or tan 0.00 0.00

X X

R R

With this value for ϕ Equation (1) reveals that

rmsrms

65.0 W123 V

cos 0.530 A cos0.00

PV

I

33. SSM REASONING The current in an RCL circuit is given by Equation 23.6,

I V Zrms rms

/ , where the impedance Z of the circuit is given by Equation 23.7 as

Z R X X 2 2( )L C

. The current is a maximum when the impedance is a minimum for

a given generator voltage. The minimum impedance occurs when the frequency is f0,

corresponding to the condition that X XL C , or 2 1 2

0 0 f L f C / ( ) . Solving for the

frequency f0, called the resonant frequency, we find that


fLC

0

1

2

Note that the resonant frequency depends on the inductance and the capacitance, but does

not depend on the resistance.

SOLUTION a. The frequency at which the current is a maximum is

fLC

0

1

2

1

2 17 0 10 10

( . –3 –6 H)( F)12.0352 Hz

b. The maximum value of the current occurs when f f0. This occurs when X X

L C , so

that Z R . Therefore, according to Equation 23.6, we have

IV

Z

V

Rrms

rms rms 155 V

10.0 15.5 A

34. REASONING The resonant frequency is given by Equation 23.10 as f LC0

1 2 / d i

and is inversely proportional to the square root of the circuit capacitance C. Therefore, to

reduce the resonant frequency, it is necessary to increase the circuit capacitance. The

equivalent capacitance CP of two capacitors in parallel is CP = C

1 + C

2 (Equation 20.18),

which is greater than either capacitance individually. Therefore, to increase the circuit

capacitance, C2 should be added in parallel with C

1.

SOLUTION The initial resonant frequency is f01

. The resonant frequency that results after

C2 is added in parallel with C

1 is f

0P. Using Equation 23.10, we can express both of these

frequencies as follows:

fLC

fLC

01

1

0

1

2

1

2

and

P

P

Here, C

P is the equivalent parallel capacitance. Dividing the expression for f

01 by the

expression for f0P

yields

f

f

LC

LC

C

C

01

0

1

1

1 2

1 2P P

P /

/

d id i

According to Equation 20.18, the equivalent capacitance is C

P = C

1 + C

2, so that this

frequency ratio becomes

f

f

C C

C

C

C

01

0

1 2

1

2

1

1P


Squaring both sides of this result and solving for C2, we find

2

01 2

0P 1

2 201

2 10P

1

7.30 kHz1 2.60 F 1 1.8 F

5.60 kHz

f C

f C

fC C

f

35. REASONING As discussed in Section 23.4, a RCL series circuit is at resonance when the

current is a maximum and the impedance Z of the circuit is a minimum. This happens when

the inductive reactance XL equals the capacitive reactance XC. When XL = XC, the

impedance of the circuit (see Equation 23.7) becomes 22

L CZ R X X R , so the

impedance is due solely to the resistance R. The average power dissipated in the circuit is 2

rms/P V R (Equation 20.15c). This relation can be used to find the power when the

variable resistor is set to another value.

SOLUTION The average power 1

P dissipated when the resistance is R1 = 175 is

21 rms 1

/P V R . Likewise, the average power 2

P dissipated when the resistance is R2 = 562

is 22 rms 2

/P V R . Solving the first equation for 2rms

V and substituting the result into the

second equation gives

2rms 1 1

22 2

2.6 W 175 0.81 W

562

V P RP

R R

36. REASONING The resonant frequency of an RCL circuit is given by

0

1

2f

LC (Equation 23.10), where L is inductance and C is the capacitance. Because

only the inductance of this circuit changes, from L1 = 7.0 mH to L2 = 1.5 mH, we obtain the

initial and final resonant frequencies from Equation 23.10:

01 02

1 2

1 1 and

2 2f f

L C L C (1)

We will solve the first of Equations (1) for the capacitance C, and substitute the result into

the second of Equations (1).

SOLUTION Squaring both sides of the first of Equations (1) and solving for C, we obtain

2

01 2 2

1 01 1

1 1 or

2 2f C

L C f L (2)


Substituting Equation (2) into the second of Equations (1) yields

02

22 2

01 1

21 1

2 12

2

fL C

Lf L

01

2

f

101

22

1

Lf

LL

L

(3)

In Equation (3) the initial resonant frequency is multiplied by the square root of the ratio of

the inductances, so that if we express the initial resonant frequency in kHz, the final

resonant frequency will also be expressed in kHz, as requested. Similarly, we do not need to

convert the inductances from millihenries to henries, since their units will cancel out. From

Equation (3), then, the final resonant frequency of the circuit is

102 01

2

7.0 mH1.3 kHz 2.8 kHz

1.5 mH

Lf f

L

37. REASONING Since the resonant frequency f0 is known, we may use Equation 23.10,

fLC

0

1

2

to find the inductance L, provided the capacitance C can be determined. The

capacitance can be found by using the definitions of capacitive and inductive reactances.

SOLUTION

a. Solving Equation 23.10 for the inductance, we have

Lf C

1

4 2

0

2 (1)

where f

0 is the resonant frequency. From Equations 23. 2 and 23.4, the capacitive and

inductive reactances are

Xf C

X f LC L

and 1

22

where f is any frequency. Solving the first of these equations for f, substituting the result

into the second equation, and solving for C yields CL

X X

L C

. Substituting this result into

Equation (1) above and solving for L gives

Lf

X X 1

2

1

2 150030 0 5 0 1 3 10

0

3

L C HzHb gb gb g. . .

b. The capacitance is

CL

X X

L C

HF

1 3 10

30 0 5 08 7 10

3

6.

. ..

b gb g


38. REASONING The inductive reactance XL is given by L 2X f L (Equation 23.4), where

f is the nonresonant frequency in hertz (Hz) and L is the inductance. The capacitive

reactance XC is given by C

1

2X

f C (Equation 23.2), where C is the capacitance. The

resonant frequency f0 is 0

1

2f

LC (23.10). From the values given for the ratio

L

C

5.36X

X

and the resonant frequency 0

225 Hzf we will determine the nonresonant

frequency f.

SOLUTION Using Equation 23.4 for XL and Equation 23.2 for XC, we can write the ratio of

the two reactances as follows:

2 2L

C

24

1/ 2

X f Lf LC

X f C

(1)

Solving Equation (1) for the nonresonant frequency f shows that

2 2L L L2

C C C

Resonantfrequency

1 14 or

24

X X Xf LC f

X X X LCLC

(2)

We note in Equation (2) that the term in parentheses is the resonant frequency f0 as given by

Equation 23.10, so that we have

L0

C

5.36 225 Hz 521 HzX

f fX

39. SSM REASONING AND SOLUTION At the resonant frequency f0, we have

C f L 1 4 2

0

2/ ( ) . We want to determine some series combination of capacitors whose

equivalent capacitance Cs is such that f f

0 03 . Thus,

FHG

IKJC

f L f L f LC

s

1

4

1

4 3

1

9

1

4

1

92

0

2 2

0

2 2

0

2 ( )

The equivalent capacitance of a series combination of capacitors is 1 1 1

1 2/ / / ... C C C

s

If we require that all the capacitors have the same capacitance C, the equivalent capacitance

is

1 1 1

C C C

n

Cs

where n is the total number of identical capacitors. Using the result above, we find that


1 1

19

C C

n

Cn

s

or = 9

Therefore, the number of additional capacitors that must be inserted in series in the circuit

so that the frequency triples is n n 1 8 .

40. REASONING

a. When a capacitor stores charge, it also stores electrical energy. The energy stored by the

capacitor can be expressed as Energy = 212

/q C , according to Equation 19.11c.

b. There is no resistance in the circuit, so no energy is lost as it shuttles back and forth

between the capacitor and the inductor. The energy removed from the capacitor when it is

completely discharged is 212

/q C . This energy is gained by the inductor. The energy

stored by an inductor is given by Energy = 212

LI (Equation 22.10), where L is the

inductance and I is the current. The maximum energy stored by the inductor is 21max2

LI ,

where Imax is the maximum current in the inductor.

SOLUTION a. The electrical energy stored in the fully charged capacitor is

2

6261 1

2 2 6

2.90 10 CEnergy = 1.17 10 J

3.60 10 F

q

C

(19.11c)

b. Since the energy stored by the capacitor is equal to the maximum energy stored by the

inductor, we can write

2

21 1max max2 2

Energy storedby inductorEnergy stored

by capacitor

or q q

LI IC LC

The maximum current in the inductor is

63

max3 6

2.90 10 C5.58 10 A

75.0 10 H 3.60 10 F

qI

LC


41. REASONING We will find the desired percentage from the ratio B A/L L . The beat

frequency that is heard is 0B 0A

f f , and the resonant frequencies are 0B B1/ 2f L C

and 0A A1/ 2f L C , according to Equation 23.10. By expressing the beat frequency in

terms of these expressions, we will be able to obtain B A/L L .

SOLUTION Using Equation 23.10 to express each resonant frequency, we find that the

beat frequency is

0B 0AB A

1 1

2 2f f

L C L C

Factoring out the term A1/ 2 L C gives

B A A0B 0A

B BA A A

A

1

21 1 11 1 1

12 2 2

2

L C L Lf f

L LL C L C L C

L C

Note that LB is greater than LA, so that A B/ 1L L is a negative quantity. Therefore, we

have written A

B

1L

L as A

B

1L

L

. Solving for A B/L L , we obtain

0B 0AA

B A

11/ 2

f fL

L L C

Remembering that 0A A1/ 2f L C , we see that this result becomes

2

0B 0A 0B 0AA A

B 0A B 0A

1 or 1f f f fL L

L f L f

Taking the reciprocal of this expression reveals that

B2 2

A0B 0A

0A

1 11.024

7.30 kHz11 630.0 kHz

L

L f f

f

Thus, the percentage increase of LB is 1.024 1.000 100% 2.4% .


42. REASONING AND SOLUTION The current in an RCL-circuit is

IV

R X X

2 2

L Cc h

Rearranging terms

(XL X

C)2 = (V/I)

2 R

2

Using Equations 23.2 and 23.4 for XC and X

L, respectively, we obtain

22

221 26.0 V2 108 149

2 0.141 A

Vf L R

f C I

Multiplying by f leads to

2f 2

L (149 )f 1/(2C) = 0

or

2f 2

(5.42 10–3

H) (149 )f 1/[(2 (0.200 10–6

F)] = 0

We can solve this quadratic equation for the frequencies. We obtain

f1

3 33 11 10 7 50 10 . .Hz and f Hz2

43. SSM REASONING The voltage across the capacitor reaches its maximum instantaneous

value when the generator voltage reaches its maximum instantaneous value. The maximum

value of the capacitor voltage first occurs one-fourth of the way, or one-quarter of a period,

through a complete cycle (see the voltage curve in Figure 23.4).

SOLUTION The period of the generator is T f 1/ 1/ (5.00 Hz) = 0.200 s . Therefore,

the least amount of time that passes before the instantaneous voltage across the capacitor

reaches its maximum value is 1

4

1

40 200 10T ( . s) = 5.00 –2 s .

44. REASONING The average power dissipated is that dissipated in the resistor and is

P I Rrms

2 , according to Equation 20.15b. We are given the current Irms

but need to find the

resistance R. Since the inductive reactance XL is known, we can find the resistance from the

impedance, which is Z R X 2

L

2 , according to Equation 23.7. Since the voltage and the

current are known, we can obtain the impedance from Equation 23.6 as Z = Vrms

/Irms

.

SOLUTION From Equation 23.7, we can determine the resistance as R Z X 2

L

2 .

With this expression for the resistance, Equation 20.15b for the power becomes


P I R I Z X rms

2

rms

2

L

22 Using Equation 23.6 to express the impedance, we obtain the following value for the

dissipated power

P I Z X IV

IX

FHG

IKJ

FHG

IKJ

rms

2

L

2

rms

2 rms

rms

L

2

A V

A W

2

2

22

21 75

115

1 7552 0 123.

..b g b g

45. REASONING AND SOLUTION We begin by calculating the impedance of the circuit

using Z = R X XL C

2 2 –c h. We have

Xf CC

Hz F

1

2

1

2 1350 4 10 1028 8

6 b gc h..

XL = 2fL = 2(1350 Hz)(5.30 10

–3 H) = 45.0

Z 16 0 45 0 28 8 22 82 2

. . . . b g b g

The current is therefore,

I = V/Z = (15.0 V)/(22.8 ) = 0.658 A

Since the circuit elements are in series, the current through each element is the same. The

voltage across each element is

VR = IR = (0.658A)(16.0 ) = 10.5V

VC = IX

C = (0.658 A)(28.8 ) = 19.0 V

VL = IX

L = (0.658 A)(45.0 ) = 29.6V

46. REASONING AND SOLUTION At very high frequencies the capacitors behave as if they

were replaced with wires that have zero resistance, while the inductors behave as if they

were cut out of the circuit. The drawings below show the circuits under this condition.

Circuit I behaves as if the two resistors are in parallel, and the equivalent resistance can be

obtained from R R R R RP P

as–1 –1 –1 / 2 . Circuit II behaves as if the two resistors are

in series, and the equivalent resistance is R R R RS 2 . In either case, the current is the

voltage divided by the resistance. Therefore, the ratio of the currents in the two circuits is


I

I

V R

V R

circuit I

circuit II

4 / /

/

2

2

b gb g

47. SSM REASONING The individual reactances are given by Equations 23.2 and 23.4,

respectively,

Capacitive reactance Xf CC

1

2

Inductive reactance X f LL 2

When the reactances are equal, we have X XC L , from which we find

1

22 12

f Cf L f LC or 4 2

The last expression may be solved for the frequency f.

SOLUTION Solving for f with L = 52 10 H and C = 76 10 F, we obtain

fLC

1

2

1

2 10 =

(52 (76 10 F)8.0 10 Hz

–6

1

–3 H)

48. REASONING Only the resistor, on average, consumes power. Therefore, the average

power delivered to the circuit is equal to the average power delivered to the resistor. The

average power is given by P I Rrms

2 (Equation 20.15b), where Irms is the rms current in the

circuit and R is the resistance. According to Equation 23.6, the current is given by

Irms

= Vrms

/Z, where Vrms

is the rms voltage of the generator and Z is the circuit impedance.

The impedance of the circuit is given by Z R X X 2 2

L Cc h (Equation 23.7). At

resonance the inductive reactance XL and the capacitive reactance X

C are equal, so that

Z = R.

R

R R

R

Circuit I High frequency

Circuit II High frequency


SOLUTION Substituting Irms

= Vrms

/Z into P I Rrms

2 , the average power delivered to the

cricuit can be written as 2

2 rmsrms

VP I R R

Z

(1)

Substituting Z = R into Equation (1) yields

2 2 22

rms rms rms 3.0 V0.098 W

92

V V VP R R

Z R R

49. REASONING The rms current in an inductor is Irms

= Vrms

/XL, according to Equation 23.3.

The inductive reactance is XL = 2 f L, according to Equation 23.4. Applying these

expressions to both generators will allow us to obtain the desired current.

SOLUTION Using Equations 23.3 and 23.4, we find that the current in an inductor is

IV

X

V

f Lrms

rms

L

rms 2

Applying this result to the two generators gives

IV

f LI

V

f L1

rms

Generator 1

2

rms

Generator 2

and 2 2

1 2

Dividing the equation for generator 2 by the equation for generator 1, we obtain

I

I

V

f L

V

f L

f

fI I

f

f

2

1

2

1

1

2

2 1

1

2

2

2

0 30

rms

rms

or A1.5 kHz

6.0 kHz0.075 A

.b g

50. REASONING To find the frequency at which the current is one-half its value at zero

frequency, we first determine the value of the current when f = 0 Hz. We note that at zero

frequency the reactive inductance is zero (XL = 0 ), since XL = 2 f L (Equation 23.4). The

current at zero frequency is 0rms

I = Vrms/R (Equation 20.14), since the inductor does not play

a role in determining the current at this frequency. When the frequency is not zero, the

current is given by rms

I = Vrms/Z (Equation 23.6), where Z is the impedance of the circuit.


These last two relations will allow us to find the frequency at which the current is one-half

its value at zero frequency.

SOLUTION We are given that 01rms rms2

I I , so

rms rms1 or Z = 2

Z 2

V VR

R

Since the impedance of the circuit is 2 2L

Z R X (Equation 23.7) and XL = 2 f L

(Equation 23.4), the relation Z = 2R becomes

22 2 2R f L R

Solving for the frequency gives

3

3

3 16 31.1 10 Hz

2 2 4.0 10 H

Rf

L

51. SSM REASONING Since we know the values of the resonant frequency of the circuit,

the capacitance, and the generator voltage, we can find the value of the inductance from

Equation 23.10, the expression for the resonant frequency. The resistance can be found from

energy considerations at resonance; the power factor is given by cos , where the phase

angle is given by Equation 23.8, tan ( – ) / X X RL C

.

SOLUTION a. Solving Equation 23.10 for the inductance L, we find that

Lf C

1

4

1

4 10 1010

2

0

2 2 (1.30 (2.94

3 2 –6

–3

Hz) 5.10 F) H

b. At resonance, f f0, and the current is a maximum. This occurs when X X

L C , so

that Z R . Thus, the average power P provided by the generator is P V Rrms

2 / , and

solving for R we find

RV

P rms

2 2(11.0 V)

25.0 W4.84

c. When the generator frequency is 2.31 kHz, the individual reactances are


Xf CC

(2.31 5

1

2

1

2 10 1013 5

3 –6 Hz)( .10 F).

X f LL

(2.31 2 2 10 2 10 42 7 3 –3 Hz)( .94 H) .

The phase angle is, from Equation 23.8,

FHG

IKJ

FHG

IKJ tan

–tan .–1 –1

X X

R

L C 42.7 –13.5

4.84

80 6

The power factor is then given by

cos cos 80.6 = 0.163

52. REASONING AND SOLUTION With only the resistor in the circuit, the power dissipated

is P1 = V

0

2/R = 1.000 W. Therefore, V

0

2 = (1.000 W) R. When the capacitor is added in

series with the resistor, the power dissipated is given by P2 = I

2V

0 cos = 0.500 W, where

cos is the power factor, with cos = R/Z2, and I

2 = V

0/Z

2. The impedance Z

2 is

2 22 C

Z R X . Substituting yields,

P2 = V

0

2R/Z

2

2 = (1.000 W) R

2/(R

2 + X

C

2) = 0.500 W

Solving for XC gives, X

C = R. When the inductor is added in series with the resistor, we

have P3 = V

0I3 cos = 0.250 W, where I

3 = V

0/Z

3 and cos = R/Z

3. The impedance Z

3 is

2 23 L

Z R X . Thus,

P3 = (1.000 W) R

2/(R

2 + X

L

2) = 0.250 W

Solving for XL, we find that X

L = R 3 . Finally, when both the inductor and capacitor are

added in series with the resistor we have

2 20

4 2 22 2L C

1.000 W0.651 W

3

V R RP

R X X R R R

Documents

ANSWERS TO FOCUS ON CONCEPTS QUESTIONS ... TO FOCUS ON CONCEPTS QUESTIONS 1. (d) According to 2 R s / (Equation 20.15c), the average power is proportional to the square of the rms