Physics 1402 Homework Solutions - Walker, Chapter 24

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Conceptual Exercises 10. The inductive reactance is given by

2LX L f Lω π= = At very high frequencies (i.e., as f → ∞ … frequencies well above resonance), LX gets very large. ( LX → ∞ ). This means that at high frequencies, the inductor looks like an open circuit! The effective resistance of the inductor becomes so large that it is as though the inductor were replaced by a break in the circuit. The reactance of the capacitor is given by

1 12CX

C f Cω π= =

At very high frequencies, 0CX → . This means that the capacitor looks like a short! The effective resistance of the cap becomes so small that it is as though the cap were just replaced by a wire. In light of the above, imagine replacing L with an open circuit and C with a short in the figure. Then no current at all would flow in the branch containing L and the top R. All the current would flow through the branch containing C and the bottom R. Furthermore, with C replaced by a short, the entire circuit would look just like the single bottom R connected across the voltage source. Therefore, by Ohm’s law, the rms current that would flow would be

( ) rmsrms high f

VIR

= ,

in which rmsV represents the rms voltage supplied by the source. At very low frequencies (as 0f → , i.e., frequencies well below resonance), 0LX → and CX → ∞ . This means that the inductor looks like a short and the cap looks like an open circuit. If you imagine replacing L with a wire and C with an open circuit (a break), there would be no current flowing in the branch containing C. All the current would flow through the branch containing L. And with L replaced by a short, the entire circuit would look like a single R (the one in the top branch) connected across the AC source. Therefore, by Ohm’s law, the rms current that would flow would be

( ) rmsrms low f

VIR

=

Comparing the two expressions for ( )rms high fI and ( )rms low fI , we see that the same current would flow.

18. (a.) The resonant frequency is given by

01

2f

LCπ= ,

which does not depend on R. Therefore, if the resistance in the LRC circuit were doubled, nothing would happen to the resonant frequency. (b.) The maximum current would decrease, assuming that the voltage supplied by the source stays the

same. The reason is as follows. First of all, the amplitude of the current is given by the “AC form” of Ohm’s law:

( )

max maxmax 22

L C

V VIZ R X X

= =+ −

,

in which I’ve used the fact that the impedance of the LRC is ( )22L CZ R X X= + − .

Physics 1402 Homework Solutions - Walker, Chapter 24

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When the LRC circuit is driven at its resonant frequency, L CX X= . This means that the denominator in the above expression is smaller than it is at any other frequency. Therefore, the amplitude of the current, maxI , is larger than it is at any other frequency. In particular, at the resonant frequency,

2 20Z R R= + = , and the amplitude of the current is

( ) maxmax res

VIR

=

The result of all of this is that the current is at its maximum amplitude at resonance. Now, if R were doubled, ( )max resI would become

( ) ( )max

max 2resVI

R= ,

which means ( )max resI would drop by a factor of two. Therefore, the maximum current in the circuit would decrease. Notice that I used the same maxV in both expressions for ( )max resI . This means I’m assuming that the voltage supplied by the source stays the same as R gets replaced by 2R. But this doesn’t always happen, in reality. No real voltage source is capable of supplying literally the same voltage between its two terminals no matter what resistance you stick across it. Real voltage sources always have some internal resistance, and this limits the current they can supply. When you connect very small load resistances across the source, you’re asking the source to supply a very large current to the load to keep the voltage across the load fixed. But at some point, because of the internal resistance of the source, the source simply isn’t capable of supplying enough current to the load to keep the voltage across it the same as it was before. Thus, when a very small load resistance is connected across the voltage source, we say that the voltage supplied by the source “sags,” meaning that it falls below what it was for very large load resistances. You will see this effect, called loading effect, in the lab.

Problems 3. We’re told that 120 VrmsV = and max 2.8 AI = . But this maxI is the amplitude of the current, not the rms current. From Ohm’s law,

rmsrms rms

rms

VV I R RI

= ⇒ =

But max

2rmsII = , so

max max

120 V2 22.8 A

2

rms rmsV VRI I

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎝ ⎠⎝ ⎠⎜ ⎟⎝ ⎠

61 R = Ω 6. (a.) The average power dissipated by the resistor is

av rms rmsP I V= but rms rmsI V R= , so

2

rmsav

VPR

=

Solving for R, I get

Physics 1402 Homework Solutions - Walker, Chapter 24

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( )222120 V

1.9 10 75 W

rms

av

VRP

= = = × Ω

(b.)

( )maxmax 2

2 120 V2 0.88 A1.9 10

rmsV VIR R

= = = =× Ω

(c.) The maximum power used by the bulb at any given time occurs when ( )V t and ( )I t are at their peak

values, maxV and maxI . At this time, the power dissipated by the bulb is

( )( )max max max 2 2 2rms rms rms rmsP I V I V I V= = =

but 75 Wrms rms avI V P= = , so: ( )max 2 75 W 150 WP = =

15. (a.) From the “AC form” of Ohm’s law for the cap

,rms C rms CV I X= So

, 3-3

0.500 V 1.16 10 1.16 k0.430 10 A

rms CC

rms

VX

I= = = × Ω = Ω

×

(b.)

1 12 2C

CX C

f C f Xπ π= ⇒ =

( )( )7

3 31 1.37 10 F 0.137 F

2 1.00 10 Hz 1.16 10 C μ

π−= = × =

× × Ω

(c.)

rmsrms

C

VIX

=

At 2.00 kHzf = ,

( )( )3 71 581

2 2.00 10 Hz 1.37 10 FCX

π −= = Ω

× ×

So

40.500 V 8.60 10 A 0.860 mA581 rmsI −= = × =

Ω

By a similar procedure, for 10.0 kHzf = , I find 4.30 mArmsI = .

19.

2

2 2 2 12CZ R X R

f Cπ⎛ ⎞

= + = + ⎜ ⎟⎝ ⎠

( )( )( )

2

26

125.5 2 60.0 Hz 95.0 10 F

Zπ −

⎡ ⎤⎢ ⎥= Ω +⎢ ⎥×⎣ ⎦

37.8 Z = Ω

Physics 1402 Homework Solutions - Walker, Chapter 24

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22. (a.)

( ) ( )( )

2 22 23

6

115 V

1 13.35 10 22 65.0 Hz 1.50 10 F

rms rmsrms

V VI

ZR

f Cππ −

= = =⎛ ⎞ ⎡ ⎤

+ ⎜ ⎟ ⎢ ⎥× Ω +⎝ ⎠ ⎢ ⎥×⎣ ⎦

23.09 10 A 30.9 mArmsI −= × = (b.)

1tan CXR

φ − ⎛ ⎞= ⎜ ⎟⎝ ⎠

1 1.63 k2CX

f Cπ= = Ω

So

1 1.63 ktanφ − Ω=3.35 kΩ

26.0⎛ ⎞

=⎜ ⎟⎝ ⎠

o

This is the angle by which the voltage lags the current.

26. (a.) The power factor is given by

cos RZ

φ =

2 2 2

2

cos1

2C

R R

R XR

f C

φ

π

= =+ ⎛ ⎞

+ ⎜ ⎟⎝ ⎠

( ) ( )( )

3

223

6

4.0 10cos 0.80

14.0 102 150 0.35 10

φ

π −

×= =⎡ ⎤⎢ ⎥× +⎢ ⎥×⎣ ⎦

(b.) If f is increased, CX will get smaller, which means Z will get smaller. Therefore, the power factor cosφ will increase. Physically, what this means is that the voltage and current get closer to being in phase, which means more power is delivered to the load.

40. (a.) The impedance of an RL circuit is given by

2 2LZ R X= +

So

( ) ( )2 27.0 15 17 Z = Ω + Ω = Ω (b.)

240 V 14 A17

rmsrms

VI

Z= = =

Ω

(c.)

cosav rms rmsP I V φ= , in which the power factor cosφ is given by

cos RZ

φ =

Physics 1402 Homework Solutions - Walker, Chapter 24

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So

( )( ) 37.0 14 A 240 V 1.5 10 W 1.5 kW17 av rms rms

RP I VZ

Ω⎛ ⎞ ⎛ ⎞= = = × =⎜ ⎟ ⎜ ⎟Ω⎝ ⎠ ⎝ ⎠

43. The impedance of the RLC circuit is given by

( )2

22 2 122L CZ R X X R f L

f Cπ

π⎡ ⎤⎛ ⎞

= + − = + −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

( ) ( )( )( )( )

223 3

6

11.50 10 2 60.0 155 102 60.0 12.5 10

Z ππ

−−

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟= × + × −⎢ ⎥⎜ ⎟×⎢ ⎥⎝ ⎠⎣ ⎦

31.51 10 1.51 kZ = × Ω = Ω 48. (a.) If you connect the voltmeter between terminals A and B, the voltmeter will measure the rms voltage

across just the inductor L. This is given by ,rms L rms LV I X=

But the rms current is given by

rmsrms

VI

Z= ,

in which rmsV is the rms voltage supplied to the whole RLC circuit by the voltage source and Z is the impedance of the entire RLC circuit. Now,

( ) ( )3 32 2 30 10 Hz 0.300 10 H 57 LX f Lπ π −= = × × = Ω

and

( )( )3 6

1 1 53 2 2 30 10 Hz 0.100 10 F

CXf Cπ π −

= = = Ω× ×

So

( ) ( ) ( )2 2 22 2.50 57 53 4.3 L CZ R X X= + − = Ω + Ω − Ω = Ω

Therefore,

6.00 V 1.4 A4.3

rmsrms

VI

Z= = =

Ω

And finally, ( )( ), 1.4 A 57 79 Vrms L rms LV I X= = Ω =

(b.) If you put the voltmeter between Terminals B and C, you’ll measure the rms voltage across just the cap. This is given by

( ) ( ), 1.4 A 53 74 Vrms C rms CV I X= = Ω =

(c.) If you put the voltmeter between Terminals A and C, you’ll measure the rms voltage across L and C. This voltage is given by ,rms LC rms LCV I Z= , in which LCZ means the impedance of the series LC combination. To get this impedance of L and C in series, think about the impedance of the entire LRC circuit:

Physics 1402 Homework Solutions - Walker, Chapter 24

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( )22L CZ R X X= + −

Now, to remove the effect of the resistor, I should set R equal to zero. Doing so gives

( )2LC L C L C L CZ Z X X X X X X= = − = − = − ,

since L CX X> . So ( ) ( ) ( ), 1.4 A 57 53 4.9 Vrms LC rms LC rms L CV I Z I X X= = − = Ω − Ω =

(d.) If you put the voltmeter between Terminals A and D, you’ll measure the rms voltage across the entire

RLC circuit. This voltage is the rms voltage supplied by the source, 6.00 VrmsV = .

54. At resonance, Z R= , so

rms rmsrms

V VI

Z R= =

This gives

120 V 67 1.8 A

rms

rms

VR

I= = = Ω

55. (a.) The current is a maximum at resonance. The frequency at which this happens is therefore the resonant

frequency:

( )( )0

3 6

1 1 49.6 Hz2 2 57.6 10 H 179 10 F

fLCπ π − −

= = =× ×

(b.) The impedance Z is a minimum at resonance. Therefore, the answer to this question is also 49.6 Hz.

57. (a.) They have the same reactance (i.e., L CX X= ) at the resonant frequency, 0f . For a 33- Fμ cap and a 33- Hm inductor, this occurs at

( ) ( )

20 3 6

1 1 1.5 10 Hz 0.15 kHz2 2 33 10 H 33 10 F

fLCπ π − −

= = = × =× ×

(b.) The resonant frequency is the frequency at which the inductor and capacitor have the same reactance.

So it’s the frequency I just found, 0.15 kHz .

78. (a.) 01

2f

LCπ=

And I want this to equal 85 MHz, which is 685 10 Hz× . So solving for C, I get

2 20

14

Cf Lπ

=

( ) ( )

1222 6 6

1 7.9 10 F 7.9 pF4 85 10 Hz 2.8 10 H

Cπ

−

−= = × =

× ×

(b.) The impedance of the RLC is

( )22L CZ R X X= + −

Physics 1402 Homework Solutions - Walker, Chapter 24

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Remember that Z is a minimum at resonance. Therefore, anything you do to change the resonant frequency makes Z larger. If C is made larger, this makes CX smaller, because CX is inversely proportional to C:

12CX

f Cπ=

When CX gets smaller, CX and LX won’t cancel at 85 MHz anymore. Therefore, in the impedance

( )22L CZ R X X= + − ,

the term in parentheses won’t be zero anymore. Because this term appears squared in the expression for the impedance, any deviation of L CX X− away from zero makes Z larger.

(c.) At resonance, the impedance is just equal to the resistance: resZ R=

Therefore, 5.0 Z = Ω

(d.) The capacitance found in Part (a.) makes C LX X= at a frequency of 85 MHz. Let us call this value of capacitance 0C and the corresponding value of the capacitive reactance ( )0CX . Then for this value of capacitance,

( )00 0

12CX

f Cπ= ,

and this equals LX at resonance: ( ) ( )0 0C LX X= Now, if C is made 1% higher than 0C , this means 01.01C C= . Because CX is inversely proportional to C, this means that the new value of CX will be:

( ) ( )0

0 0 0 0 0

1 1 1 1 12 2 1.01 1.01 2 1.01C CX X

f C f C f Cπ π π⎡ ⎤⎛ ⎞ ⎛ ⎞= = = =⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦

Plugging this into the formula for the impedance, I get:

( ) ( ) ( )2

22 20 0

11.01L C C CZ R X X Z R X X⎡ ⎤⎛ ⎞= + − = = + − ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

,

in which I’ve recognized that ( )0L CX X= because the inductance and the frequency didn’t change (only the capacitance). Simplifying this expression a little, I get

( ) ( )2 2

22 20 0

1 0.0111.01 1.01C CZ R X Z R X⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎡ ⎤= + − = = +⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎣ ⎦⎝ ⎠ ⎝ ⎠⎣ ⎦

But the capacitive reactance at resonance is equal to ( ) 00

2LX f Lπ= , so

[ ]2

220

0.01 21.01

Z R f Lπ⎛ ⎞= + ⎜ ⎟⎝ ⎠

( ) ( )( )2

22 6 60.015.0 2 85 10 Hz 2.8 10 H1.01

Z π −⎛ ⎞ ⎡ ⎤= Ω + × ×⎜ ⎟ ⎣ ⎦⎝ ⎠

16 Z = Ω