Chapter 35 by John Smith

AC CIRCUITS

35.1. Model: A phasor is a vector that rotates counterclockwise around the origin at angular frequency w. Solve: (a) Refemng to the phasor in Figure Ex35.1, the phase angle is U? = 180'

n rad - 30" = 150° x -= 2.618 rad 180"

2*618rad = 175 rad / s 1 5 ~ 1 s0 ~ ~

(b) The instantaneous value of the emf is & = €, cosw = (10 V)cos(2.618 rad) = -8.66 V

Assess: Be careful to change your calculator to the radian mode to work with the trigonometric functions.

35.2. Model: A phasor is a vector that rotates counterclockwise around the origin at angular frequency w. Solve: (a) Refemng to the phasor in Figure Ex35.2, the phase angle is

(b) From Figure Ex35.2, &=€ocosw*€o

& cos on

=-=

cos(~t 1 4 rad)

= 70.7 v

35.3. Model: A phasor is a vector that rotates counterclockwise around the origin at angular velocity w. Solve: The emf is E=&,coswt = (50 V)cos(2a x 111 r a d s x 3.0 x

s) = (50 V)cos(2.092 rad) = (50 V)cos120°

Phasor at f = 3.0 rns

35.4. Model:

A phasor is a vector that rotates counterclockwise around the origin at angular velocity w. Solve: The instantaneous emf is given by the equation

&= (170 V) COS[(?I~X 60 Hz)t] At i = 60 ms, E= (170 V) cos (22.619 rad). An angle of 22.619 rad corresponds to 3.60 periods, which implies that the phasor makes an angle of (0.60)2a rad or (0.60)(360")= 2 16" in its fourth cycle.

35-1

35-2

Chapter 35

P h a h at f=60ms

35.5. Visualize: Please refer to Figure 35.4 for an AC resistor circuit. Solve: (a) For a circuit with a single resistor, the peak current is

(b) The peak current is the same as in part (a) because the current is independent of frequency.

35.6. Model: Current and voltage phasors are vectors that rotate counterclockwise around the origin at angular frequency w. Visualize: Please refer to Figure 35.6. Solve: (a) From the figure we note that V, = 10 V and I, = 0.50 A. Using Ohm's law,

(b) The frequency is f

0 1 =-=-=-28 T

1 0.04 s

- 25 Hz

i, = IRCOS@ = (0.50 A)cos[2~(25H z ) ~ ]

vR = VRCOSW= (10 V)cos[2~(25Hz)t]

For both the voltage and the current at t = 15 ms, the phase angle is

ux = 2d25 Hz)(15 ms) = 2N0.375) rad = 135"

' ql=;.50A

That is, the current and voltage phasors will make an angle of 135" with the starting t = 0 s position. I

S='O"

I I

v \ 0

Assess: Ohm's law applies to both the instantaneous and peak currents and voltages. For a resistor, the current and voltage are in phase.

AC Circuits

35-3

35.7. Visualize: Figure 35.7 shows a simple one-capacitor circuit. Solve: (a) The capacitive reactance at o= 2@= 2n(lOO Hz)= 628.3 rad/s is 1 1 = 5305 R WC (628.3 rad / s)(0.30 x 10" F)

x,=-=

10.0 v ,,=E= = 1 . 8 8 ~ 1 0 -A=1.88mA ~ X, 5 . 3 0 5 ~ 1 0C2~

(b) The capacitive reactance at o =2@100 kHz) = 628,300 rad/s is 1 1 x, =-= = 5.305 R WC (6.283 x lo5 rad / s)(0.30 x lo4 F)

Assess: Using reactance is just like using resistance in Ohm's law. Because Xc = w - ' , X c decreases' with an increase in w , as observed above.

35.8. Solve: (a) For a simple one-capacitor circuit,

When the frequency is doubled, the new current is

I; = W'CV, = (2W)CVC= 2(WCv,) = 21, = 20.0 mA (b) Likewise, when the voltage is doubled, the current doubles to 20.0 mA. (c) When the frequency is halved and the emf is doubled, the current remains the same at 10.0 mA.

35.9. Visualize: Figure 35.7 shows a simple one-capacitor circuit. Solve: (a) From Equation 35.1 I , Ic=---=&Vc=2~CV,* vc - vc xc

lld

50 x 1 0 - ~A 2+.0 v)(20x

= 79.6 kHz

(b) The AC current through a capacitor leads the capacitor voltage by 90" or n/2 rad. For a simple one-capacitor circuit i, = I, cos(wt + +n).For i, = I,, (a+ +n)must be equal to 2nn, where n = 1,2, ... This means 3n 7n lln a=---

2 ' 2 ' 2 '..' At these values of 02, v, = V, cos(u#) = 0 V. That is, i, is maximum when v, = 0 V.

35.10. Solve: From Equation 35.1 1,

35.11. Solve: The peak current through the capacitor is I, =-vc

=,xc ---"c

X C

= ~ . O X 1 0 - ~A

=125Q

Assess: Using reactance is just like using resistance in Ohm's law. This, however, does not apply to instantaneous values of current and voltage.

35.12. Model: The current and voltage of a resistor are in phase, but the capacitor current leads the capacitor voltage by 90". Solve: For an RC circuit, the peak voltages are related through Equation 35.12. We have

E; = V,'

+ Vc'

V, =

d m

= J(lO.0 V)' - (6.0 V)? = 8.0 V

35-4

Chapter 35

35.13. Model: The current and voltage of a resistor are in phase, but the capacitor current leads the capacitor voltage by 90". Viualize: Please refer to Figure Ex35.13. Solve: From Equations 35.13 and 35.14, the peak voltages are V, = IR and Vc = E,, where

=-=-=

1 2lcfc

1 = 199 Q 2n(1x104 H Z ) ( ~ O X I OF) ~

The peak current is

-4

10 v

Ell

= ,/(199 Q)'

+ (150 Q)'

= 0.0401 A

Thus, V, = (0.0401 A)(150 Q) = 6.02 V and Vc = rX,= (0.0401 A)(199 Q) = 7.99 V.

35.14. Model: The crossover frequency occurs for a series RC circuit when V, = V,. Visualize: Please refer to Figure 35.13b for the high-pass filter circuit. Solve: From Equation 35.15,

=-ac=-= 1

wc=2jq

1 2zRL

1 = 1.59 x 10" F = 1.59 pF 2n(lOO SZ)(lOOO Hz)

Assess: The output for a high-pass filter is across the resistor.

35.15. Model: The crossover frequency occurs for a series RC circuit when V, = Vc. Visualize: Please refer to Figure 35.13a for the low-pass filter circuit. Solve: From Equation 35.15, 1 =-*c=--

w,=2jq

1 1 = 1.59 x lod F = 1.59 PF 2xRf, 2 4 1 0 0 a)(lOOO Hz)

Assess: The output for a low-pass filter is across the capacitor.

35.16. Model: In the case of a series RC circuit, the crossover frequency occurs when V , = V,. Visualize: Please refer to Figure 35.13a for the low-pass filter circuit. Solve: From Equation 35.15, 1

w, = 6 2 8 0 r a d l s = RC Let the crossover frequency be w'if the capacitance is doubled, that is,

35.17. Visualize: Figure 35.14 shows a simple one-inductor circuit. Solve: (a) The peak current through the inductor is

=v,=v,=---

10.0 v

L "

OIL. 2@

= 0.796 A

2 n ( 1 0 0 H z ) ( 2 0 ~ 1 0 -H) ~

(b) At a frequency of 100 kHz instead of 100 Hz as in part (a), the reactance will increase by a factor of lo00 and thus the current will decrease by a factor of 1000. Thus, IL = 0.796 A.

35.18. Solve: (a) For a simple one-inductor circuit,

If the frequency is doubled, the new current will be

AC Circuits

35-5

(b) If the voltage is doubled, the current will double to 20.0 mA. (c) If the voltage .is doubled and the frequency is halved, the current will quadruple to 40.0 mA.

35.19. Model: The AC current through an inductor lags the inductor voltage by 90’ Solve:

(a) From Equation 35.21,

I,=50mA =

L V ,* f = x, 2@

2n(50 x

5.0 V = 3.18 x lo4 HZ A)(500 x 10“ H)

(b) The current and voltage for a simple one-inductor circuit are iL = I, COS(OX - +n) v, = v, casu# For iL = Z, m - + z must be equal to 2nn, where n = 0, 1, 2,

...

This means

=(2nz++z).Thus, the

instantaneous value of the emf at the instant when iL= IL, is v, = 0 V.

35.20. Solve: For a simple one-inductor circuit, IL,2LV,*L XL 2@

&Vms

=-=--VL

2nf I,

2nf I ,

f i ( 6 . 0 V) = 1.39 x 21c(15 x lo3 Hz)(65 x IO” A)

H = 1.39 mH

35.21. Solve: For a simple one-inductor circuit, I

---j

L -

1.0 v X , = L = 80 i2 I, 1 2 . 5 ~ 1 0 ”A

Assess: Using reactance is just like using resistance in Ohm’s law. This, however, does not apply to instantaneous values of current and voltage.

35.22. Solve: When a capacitive reactance X, and an inductive reactance X, become equal,

-* w = J- 1 wc LC

WL = 1

*f=-

= 159 kHz

2n J(1.0~10” H)(I.OxIO” F)

The reactance at this frequency is

X, = X, = OX= 27z( 159 kHz)( 1.O x 10-6H) = 1 .O R

35.23. Model: At the resonance frequency, the current in a series RLC circuit is a maximum. The resistor does not affect the resonance frequency. Solve: From Equation 35.30,

1 4n’(1000 Hz)’(20

x IOe3 H)

= 1.27 x 10“ F = 1.27 pF

35.24. Model: At the resonance frequency, the current in the series RLC circuit is a maximum. The resistor does not affect the resonance frequency. Solve: From Equation 35.30,

3Lz-z 1 4n’f ‘C

35.25. Solve:

1 = 10.1 mH 4z2(1000 Hz)’(2.5x 10“ F)

(a) When the resistance is doubled, the resonance frequency stays the same because f is independent of R . Hence, f = 200 kHz. (b) From Equation 35.30, 1 1 f =-2rr LE When the capacitor value is doubled, 1- 1 f - 200 kHz - 141 lcHz f’= =21cJqq 4 5 - 7 -

35-6

Chapter 35

35.26. Solve: (a) When the resistor value is doubled, the frequencyf is unchanged because f does not depend on R. H e n s f = 200 kHz. (a)From Equation 35.30, f=-

'si-

2K LC When the capacitor value is doubled and the inductor value is halved, then

35.27. Model: We have an RLC circuit. Visualize: Please refer to Figure Ex35.27.

Solve: (a) The inductive reactance and capacitive reactance are X , = OIL= 2 ~ ( 1 4x lo3 Hz)(lO x

H) = 879.6 R

xc =-=1

2rc(14 x lo3 Hz)(lOx

= 1136.8 s2

Because X , > X,, the circuit is capacitive and the current leads the voltage. From Equation 35.27,

Thus, the emf voltage lags the current by 27.2" at the frequency of 14 kHz. (b) At a frequency of 18 kHz, X , = 1130.9 R and X, = 884.2 R . Because X , > X,, the circuit is inductive and the emf voltage leads the current by an angle 9 = +26.3".

35.28. Visualize: Assume a simple one-resistor circuit. Solve: From Equation 35.39,

Using the equation again,

= J(lO.0 W)(50.0 R) = 22.4 V

35.29. Visualize: Assume a simple one-resistor circuit. Solve: An electrical device labeled 1500 W is designed to dissipate an average 1500 W at V, = 120 V. We can use Equation 35.39 to find that

Assess:

Calculations with rms values are just like the calculations for DC circuits.

35.30. Visualize: Assume a simple one-resistor circuit. Solve: An electrical power outlet in the United States is 120 V160 Hz. This is E,.

We can use Equation 35.39 to find

35.31. Model: The energy supplied by the emf source to the RLC circuit is dissipated by the resistor. Because of the phase difference between the current and the emf, the energy dissipated is PR = I,,&, COS@ = I,,V,,

(Equation 35.47). Solve: From Equation 35.28,

= &,cos@

Using Ohm's law, R=%!L=,,- E 1,s

COS@ Inns

= E,

- (120 ~

~ 2.4 A

cos@ 8 = 43.5 7 )12

AC Circuits

35-1

35.32. Model: The energy supplied by the emf source to the RLC circuit is dissipated by the resistor. Because of the phase difference between the current and the emf, the energy dissipated is PR = Ims&ms cos4 = Im,Vm (Equation 35.47).

Soive: The power supplied by the source and dissipated by the resistor is Psowe= PR= Pmaxcos2@ where P,, = 1

E, is the maximum possible power. For an RLC circuit, I,, = &,/R, so

Thus the power factor is

35.33. Model: The motor has inductance, otherwise the power factor would be 1. Treat the circuit as a series RLC circuit without the capacitor (X, = 0 Q). Solve: The average power is Pmme= ImSEmsCOS$ = (3.5 A)(120 V)cos2O0 = 395 W

35.34. Solve: (a) From Equation 35.14,

(b) At this frequency, V,ZIR=~ ~2

&QR + (w~apc)-'

'QR - -4? E, R2 + R2/3 2

w, = - =1 6 2 8 0 r a d / s ~ o w c a p =&a, =1.09x1O4radIs RC

35.35. Solve: (a) From Equation 35.14,

(b) At this frequency,

1 RC

o,=-=

6280 rad I s

w,, = oc/& = 3630 rad I s

35.36. Visualize: Please refer to Figure P35.36. Solve: (a) The voltage across the resistor is V, = IR =

35-8

Chapter 35

The values of V, at various values offare in the following table.

(b)

f (Hz)

VR (VI

100 300 1000 3000 10,000

0.994 2.87 7.07 9.48 9.95

VR (v) 10 -

Assess:

f (&)

The circuit acts like a high-pass filter when the output voltage is taken across the resistor.

3537. Visualize: Please refer to Figure P35.37. Solve: (a) The voltage across the capacitor is

The values of V, at a few frequencies are in the following table.

v c (VI 9.95 9.57 7.05 3.15 0.990

Z )

1 3 10 30 100

o i

Assess:

f(kH2)

100

For the voltage across the capacitor, the circuit is il low-pass filter.

AC Circuits

35.38.

35-9

Visualize: capacitors

Iu I

Solve: For a circuit with parallel capacitors, C, = C , + C,. The current in the circuit is

m e n these two capacitors are connected in series,

=e ';

+ C;'. Using the same formula for the current,

We have two simultaneous equations for C , and C,. The solutions are C , = 3.15 pF and C, = 5.53 pF.

35.39. Visualize: Please refer to Figure 35.13. Solve: For an RC filter, I = Eo/J= and

At the crossover frequency o = w, = 1/RC and thus l/w,C = R. Hence,

35.40. Visualize: Figure 35.18 defines the phase angle 4. Solve: The phase angle is

A capacitor-only circuit has no resistor (hence R = 0 Q) and no inductor (hence X , = 0

4 = tan-'(?)

0-x,

= tan-'(-.)

a).Thus the phase angle is

= --rad

That is, the current leads the voltage by d 2 rad or 90". That is exactly the expected behavior for a capacitor circuit.

35.41. Visualize: Please refer to Figure 35.13a for a low-pass RC filter. Solve: (a) From Equation 35.15, the crossover frequency is 1 f,=-= 2nRC

(b) The capacitor voltage in an RC circuit is

2z(159 Q)(lOO x lo6 F)

= 10.0 Hz

35-10

Chapter 35

where, in the last step, we factored capacitor voltage as

l? from the

square root. Using the definition w, = l/RC, we can write the

&,w, l o

EOCf,/f)

Using Eo = 5.0 V, we find

f +.L

vc (V)

4.47 3.53 2.24

fc 2fc

Assess: As expected of a low-pass filter, the voltage decreases with increasing frequency.

35.42. Visualize: Please refer to Figure 35.13b for a high-pass RC filter. Solve: (a) From Equation 35.15, the crossover frequency is f, =--

2nRC

2n(100 Q)(1.59 x lo6 F)

= 1000 HZ

(b) The resistor voltage in an RC circuit is

where, in the last step, we factored capacitor voltage as

l? from

the square root. Using the definition w, = 1/RC, we can write the

EO Using E, = 5.0 V, we find

V R (VI

+.L

2.24 3.53 4.47

fc 2.L

Assess: As expected of a high-pass filter, the voltage increases with increasing frequency.

35.43. Visualize:

f = 200

Solve: (a) The figure shows that the equivalent capacitance of the three capacitors is 2 pF. The capacitive reactance and peak current are 1 1 x, =-2nfC - 2n(200 Hz)(2 x IO6 I=

5 =? -!! = 0.02513 A x, 397.9Q

= 397.9 n

= 25.1 mA

(b)All the current passes through the 3 pF capacitor. Thus the peak voltage across the 3 pF capacitor is 1 V,, = ZX, = (25.13 x A) 2 ~ ( 2 0 0H2)(3 x 10" F) = 6.67 V

AC Circuits

1-:r/300a fi

35.44. Model: 20 V is the rms voltage. vialize: 200 R

35-11

200 R

20 VI200 Hz

Solve: (a) Using the law of series and parallel resistances, the figure indicates an equivalent resistance of 400 R in the circuit. Thus,

Ims P,,,,

=_v-s=-- 20 v R

400Q

- 0.050 A

= (0.050 A)(20 V) = 1.0 W

= Z,,V,

(b) The power dissipated by the 200 i2 resistor and its voltage drop are

e,,

= IL,(200 R)= (0.050 A)2(200R)= 0.50 W

This means the voltage across the 600 R and the 300 R resistors is 20 V - 10 V = 10 V. Thus, the power dissipated by the 600 C2 resistor is

Likewise, P,,,

= 0.333 W

35.45. Solve: From Equation 3 1.36, the voltage of a discharging capacitor is

The crossover frequency for a low-pass circuit is 1 2 ~ ( 3 . 6 0 7x

= 44.1 Hz

35.46. Model: The AC current through a capacitor leads the capacitor voltage by 90". On the other hand, the current and the voltage are in phase for a resistor. Visualize: Parallel circuit components have the same potential difference. The voltage phasor of both the resistor and capacitor match the phasor of the emf.

Please refer to Figure P35.46.

35- 12

Chapter35

Solve: (a) Because we have a parallel RC circuit, the voltage across the resistor and the capacitor is the same. The current phasor I , is therefore along the same direction as the voltage phasor Eo, and the current phasor IC is ahead of the voltage phasor Eo by 90". The peak currents in the resistor and capacitor are

(b) From the figure, we see that the phasors I, and I, are perpendicular to each other. We can combine them and find the phasor for the peak emf current I as follows:

35.47.

Model: While the AC current through an inductor lags the inductor voltage by 90", the current and the voltage are in phase for a resistor. Visualize: Series elements have the same current, so we start with a common current phasor I for the inductor and resistor,

"\ C

Please refer to Figure P35.37. Solve: Because we have a series RL circuit, the current through the resistor and the inductor is the same. The voltage phasor V, is along the same direction as the current phasor I . The voltage phasor V, is ahead of the current phasor by 90".

(a) From the phasors in the figure, Eo =

d-.

Noting that V, = IwL and V, = IR, Eo = I

V, = $--

m and

&OR R2 + w2L2

R2+ w2L2

(b) As w + 0 rads, V, + EoR/R= &, and as W + -, V, + 0 V. (c) The LR circuit will be a low-pass filter, if the output is taken from the resistor. This is because V, is maximum when w is low and goes to zero when w becomes large. (d) At the cross-over frequency, V, = V, . Hence,

R Iw,L= I R ~ w =, L

-/,

= (100

a)' +

2n(60 Hz)(O.lO H) -

AC Circuits

35-13

The peak current I is

(b) The phase angle is

This is an inductive circuit because (c) The average power loss is

4 is positive.

PR = I,&,

=(120 cos(6.4") = 143 W 100 Q

cos4 = -cos0 R

35.49. Visualize: The circuit looks like the one in Figure 35.17. Solve: (a) The impedance of the circuit is

Z =,

= -,/(lo0 Q)'

2 4 6 0 Hz)(0.15 H) -

w =

+ (56.55 R - 88.42 Q)'

1 2 ~ ( 6 Hz)(30 0 x lo4 F)

= 104.96 R.

The peak current is

I = a&. = (120V)fi = 1.617A = 1.62A Z 104.96 (b) The phase angle is

4 = tan-'(*) Because

x -x,

56.55 R - 88.42 Q ) loo R = -17.7'

= tan-'(

4 is negative, this is a capacitive circuit.

PR= I,,E,,

cosq = -COSij

cos(-17.7") 100 R

= (120

= 137 W

35.50. Visualize: Please refer to Figure P35.50. Solve: (a) The resonance frequency of the circuit is

w=-- 1

f i - J(10

1 = 3160 rad / s H)(lO x 10" F)

w *f == 503 Hz 2n

(b) At resonance, X , = X,. So, the peak values are

, = & o = = o-v1.0A - *VR= R

10R

IR= (1.0 A)(10 Q ) = 1O.OV

a V, = I X , = I w L = (1.0 AX3160 rad/s)(lO x ( c ) The instantaneous voltages must satisfy vR + vc

+ v,

H) = 31.6 V

6.But v, and v, are out of phase at resonance and cancel. Consequently, it is entirely possible for their peak values V, and V, to exceed E,. VR + V, + V, = &o is not a =

requirement of an AC circuit.

35.51. Visualize: Please refer to Figure P35.5 1. Solve: (a) The resonance frequency of the circuit is

w=-- 1

f i - J(l.0

x lo-? H)(l.Ox 10" F)

= 3 . 1 6 ~ 1 0 "r a d / s * f = - =

w 2K

5 . 0 3 ~ 1 0Hz ~

35-14

Chupfer 35

(b) At resonance X , = X,. So, the peak values are

E 1ov I=m=-= R 10R

* v, = m, =

1.0A * V R = I R = ( l . O A ) ( l O a ) = 1O.OV

(2) =

1.0 A = 31.6 V ( 3 . 1 6 ~ 1 0rad/s)(1.0x106 ~ F)

(c) The instantaneous voltages must satisty v, + v, + v, = E. But v, and v, are out of phase at resonance and cancel. Consequently, it is entirely possible for their peak values V , and V, to exceed E,,. V, + V, + V, = &, is not a requirement of an AC circuit.

35.52.

Visualize: Please refer to Figure P35.52. When the frequency is very small, X , = l/oC becomes very

large and X , = oL becomes very small. So, the branch in the circuit with the capacitor has a very large impedance and most of the current flows through the branch with the inductor. When the frequency is very large, the reverse is true and most of the current flows through the branch with the capacitor. Solve: (a) When the frequency is very small, the branch with the inductor has a very small X,, so Z z R . The current supplied by the emf is 10 v I, = -= 0.10 A loo a

(b) When the frequency is very large, the branch with the capacitor has a very small X,, so Z z R . The current supplied by the emf is

I IOlS =--lo" - 0.20 A 50

,/-

= J(50

= (50 Q)'

2n(3 x lo3 Hz)(3.3 x 10" H) -

1 2 ~ ( x3 io3 H Z ) ( ~ x~ 1O0 - ~F)

R)*+ (62.20 R - 110.52 R)? = 69.53 i2

The peak current is

The phase angle is

(b) For 4000 Hz, Z = 50.0 R, I = 0.100 A, and Q = 0". (c) For 5000 Hz, Z = 62.42 R. I = 0.080 A, and Q = +36.8" The following table summarizes the results. f =3000 HZ

z (R) I (A)

69.53 0.072 44.0"

f = 4000 Hz

f =5000 Hz

50.0 0.100 0"

62.42 0.080 36.8"

35-15

AC Circuits

-/,

= (50 R)'

= d(50 R)'

2z(5.0 x lo3 Hz)(3.3 x

H) - 2+0

1 x io3 H Z ) ( ~ x~ O

+ (37.36 R)' = 62.42 Q

Since E = E, COSUB, E = &o implies that

wt = 0. From

Equation 35.22, the instantaneous current is

Ell i = I cos(m - @)= -COS(W - @)= -cos(0" - 36.8") = 0.064 A Z 62.42 R (b) &= 0 V and E decreasing implies that w = Sarad. As before, .

1=-

5.0 v cOs(90" - 36.8") = 0.048 A 62.42 R

35.55. Visualize: The circuit looks like the one in Figure 35.17. Solve: (a) The instantaneous current is i = I cos(un - @).The phase angle is

Since i = 1cos(m - @), i = I implies that w - @ = Orad. That is, ot = Cp = -44.0" . Thus,

E = &, COS^ = (5.0 V ) ~ o ~ ( - 4 4 . 0 "=) 3.60 V (b) i = 0 A and is decreasing implies that w - Cp = +arad. That is, w = + K + @ = 90" - 4 . 0 " = 46.0". Thus, & = (5.0 V) cos(46.0") = 3.47 V . (c) i = -I implies m - Cp = arad. That is, m = T + @ = 180" - 44" = 136". Thus, I =(5.0 V)cos136" = -3.60 V.

35.56. Model: The average power is the total energy dissipated per second. Visualize: The circuit looks like the one in Figure 35.17. Solve: (a) The resonance frequency is w, = At w = +a,, the inductive and capacitive reactance are

x, - x,

= --3 J T = --3 / i 5 x K 4 7 4 3 Q 2 c 2 1.0~10F -~

* tan@= 4743 * - 47.43 rad 100 R

-4

*z=

= J(4743

@ = -88.8"

a>'+ (loo R)'

= 4744 R

From Equation 35.47, the average power supplied to the circuit is

10 v = l,,â&#x201A;Ź,, cos@=(-)(lo 4744 R

V)cos(-88.8") = 4.4 x lo4 W

(b) At OJ = 0,. X , = X,. So, Z = R and @ = 0. The average power supplied by the emf is

35-16

Chapter 35

(e) At w = 2%, the inductive and capacitive reactance are

As in part (a), 4 = +88.8" and Z = 4744 R.The average power is Pmme= 4.4 x lo4 W. Assess: The power dissipated by a RLC circuit falls off dramatically on either side of resonance.

35.57. Solve: From Equation 35.25, the impedance is 2 = JR2 inductive reactance and the capacitive reactance can be written

35.58. Solve:

+ ( X , - Xc)* . The difference

between the

(a) The peak current in a series RLC circuit is

The maximum current is ,,Z

z = ,IRZ+ (x, - X J *

= E 0 / R and it occurs only at resonance because this is when the impedance

is the smallest. Using this expression for Z,

=--

- cos@

Thus, I = I,, cos@. (b) From Equation 35.44, the average power is

pavg = +ZE, cos4 = +(Iwhere P,,

35.59.

cos4)&, cos4 = (+I-â&#x201A;Ź,)cOs2

4 = P,,

cos2@

is the maximum power the source can deliver to the circuit.

Solve: (a) The resonance frequency is

m -2?q0*C=--

w0=--

1 1 = 11.64 pF 4r2f,2L 4r2(104.3 x lo6 H ~ ) ~ ( 0 . 2 0pH) 0

(b) The current produced by the out-of-tune radio station is 0.10% of the resonance current. Therefore,

1 2~(103.9MHz)(0.20 pH) - 2r(103.9 MHz)(l1.6 pF) = 1.5 x 1 0 - ~R Assess: The impedance at resonance is Z = R because X , = X,.

35.60. Solve: (a) The resonance frequency is f, =57MHz =

1 1 *L=-= 2nfi 4n-fic

1 4n2(57x lo6 Hz)'(l6 x IO-" F)

= 0.487 jfH

AC Circuits

35-17

Forf= 60 MHz, 2n(60 x lo6 Hz)(0.487 pH)= -[183.70 1

1 2a(60 x lo6 Hz)(16 x lo-'' F)

R - 165.79 R]= 10.3 R

Forf= 54 MHz, R = 10.9 R. The minimum possible value of the circuit resistance is thus 10.9 R.

35.61. Model: The filament in a light bulb acts as a resistor. Visualize: Please refer to Figure P35.6 1. Solve: A bulb labeled 40 W is designed to dissipate an average of 40 W of power at a voltage of Vm = 120 V. From Equation 35.39, the resistance of the 40 W light bulb is

R40---

IS - (120V)' - 360 R

40W

Likewise, RW = 240 R and R,, = 144 R . The rms current through the 40 W and 60 W bulbs, which are in series, is

I = I =-= 600R The rms voltage across the light bulbs in series are

120 v = 0.20 A 600R

V, = I d a = (0.20 A)(360 Q) = 72 V

V,, = Idso = (0.20 A)(240 Q) = 48 V

The voltage across the 100 W light bulb is V,, = 120 V. The powers dissipated in the light bulbs in series are = V J , = (72 V)(0.20 A) = 14.4 W

P, = VJ,

= (48 V)(0.20 A) = 9.6 W

The power dissipated in the 100 W bulb is

35.62. Model: The wires are resistors as well as conductors. Solve: (a) From Equation 35.40, P, = (I,, Ern), + (Irrn E,)?

+ (ImErn)) 3 450 MW = 31-

&- = 31m( 120 V)

(b) Using the same equation, 450 MW = 31m&ms= 3Im(5O0 x 10' V)

I, =

450 x lo6 W = 300 A 3(500 x 10' V)

(c) Copper may be a good conductor, but it still has some resistance. Even a thick copper cable cannot c q a 1.25 million amp current. It would melt! A step up to much higher voltages allows the power to be transmitted at much lower currents.

35.63. Model: Assume the motor is a series RLC circuit. Solve: (a) From Equation 35.44, the average power is f&, = I,,&, cos@-

cos@.Thus, the power factor is

W pI,, ,,,,- (1 20800 = 0.833 ,&,, V)(8.0 A)

35- 18

Chapter 35

(b) The resistance voltage is Vm = &m cos$ = (120 V)(0.833) = 100 V. (c) Using Ohm's law, the resistance of the motor is

100 v = 12.5 Q 8.0A

(a)From the results of part (a) and from Equation 35.27, the phase angle is 4 = cos-'(O.833) = 33.6" = tan-' 3X,

- X , = (12.5 Q)tan33.6" = 8.292 Q

The phase angle becomes zero and the power factor becomes 1.0 by increasing the capacitive reactance by 8.292 Q. The capacitor with this reactance is 1 = 3.20 x lo4 F = 320 pF C= 2n(60 Hz)(8.292 SZ)

35.64.

Visualize:

€0

Solve: (a) The three phasors are shown in the figure. The second phasor has a phase 120" ahead of the first and the third phasor has a phase that is 120" behind the first. These phasors are labeled as (Eo)',(&o)2, and (&o)3. (b) As shown in the figure, the three vectors (or phasors) add to zero. This means that the sum of the three phasors is zero. (c) In the figure above, we can use the law of cosines to find &; = J&; + €;

- 2E0€, cos$ = J&; + &; - 2&,2cos(120") = &&o

Thus the rms value of the difference between two phases is

&LS= &&m,

= &(I20 V) = 208 V

35.65. Model: The current and voltage of a resistor are in phase. For a capacitor, the current leads the voltage by 90". Visualize: This is the phasor diagram for a RC circuit in which the current leads the emf voltage by 45". I

' Eo

45"

AC Circuits

35-19

Solve: We have two tasks. First, to get the phase right. Second, to get the output voltage right. In the RC circuit above, the resistor voltage is in phase with the current. If we chose a capacitor and resistor such that V, = Vc, then the current and the resistor voltage (the output) lead the emf (the input) by exactly 45".Since V, = IR and V, = ZX, the phase angle will be 45" if R = X , = I/&. This gets the phase right, but is the output voltage right? Under these conditions, the rms voltage across the resistor is

This is too much voltage. We want a circuit in which (VR)m = 0.5 Em. But we can reduce the voltage by splitting the resistor R into two series resistors whose total resistance is R.

T 11.ov

t------t

R, = 18.7 C?

6.0 V

Because the current is the same through both resistors, -=-;=V, + Vz

V, V,

R, R, + R2

- -R'

R2 = -R V,

0.500â&#x201A;Ź,,,,, R = 0.707R 0.707â&#x201A;Ź,,

That is, if we split resistor R into R, = 0.293R and R2= 0.707R, and take the output from R,, then the total resistance will cause the phase angle to be 45" and the voltage across R, will be half the input voltage. There are no unique values for R and C. Any values that obey these relationships will work. But to be specific, suppose we choose C = 100 pF, a very typical capacitance value. Then

R=-=1

1 = 26.5 ( 2 x x 60 Hz)( 100 x 10" F)

R * I?,

= 0.293R = 7.8 S2 and R2 = 0.707R = 18.7 S2

Hence our circuit is as shown above.

35.66. Model: Assume that the circuit is a series RLC circuit. Solve: (a) From Equation 35.44, P ,, = I,,,,,&,, COS^. Hence, 6.0 x IO6 W = Im(15,OOO V)(0.90) (b)

Im = 444 A

The figure shows the nature of R, X,, X,, and Z as a phasor diagram. The power factor is cos@= 0.9. The total = ( 15,000 V)/(444A) = 33.8 R.You can see from the right triangle in circuit impedance is found from Z = &,,/I,, the diagram that

XL.- X, = Zsin@= Z

x = (33.8 R),/1-(o.s,z

= 14.7 R

35-20

Chapter 35

You can bring the power factor up to 1 .O by reducing X, - X, to zero. To do this, you can increase X, by 14.7 il. Since X, = l/wC, the extra capacitance needed to raise the power factor is

1 c=(27r x 60 Hz)(14.7 =1.8Ox1O4 F = R)

180pF

(c) When the power factor is 1.0, the power delivered is P = (&m)z/R. From the figure above, we see that the resistance is R = &os@= (0.9)(33.8Q) = 30.4 R. Thus

35.67. Solve:

(a) The peak inductor voltage in a series RLC circuit is

For VL to be maximum, dV,/dw = 0. Taking the derivative and dividing by the constant factors,

where O, = @ and the maximizing frequency is q. (b) The voltage across the inductor is

w = w,,= l 1

/ Z , Z = R = 1.0 Q . We get

VL =-m&=Z

€0

J-

L =-1OV l.0pH C

l.0pF

The maximizing frequency q is

The impedance at this frequency is

Z = , / R 2+ (wLL- ~ / W , C ) ~ ( 1 . 4 1 4 ~ 1 0rad/s)(l.OpH)~

1 = 1.225 R (1.414x 10" rad / s)(1.0 pF)

The voltage is

v, = -w,L €0

1.225 R

= 'O.OV ( 1 . 4 1 4 ~ 1 0rad/s)(l.OpH)=11.55 ~ V

35.68. Solve: (a) From Equation 35.44, the average power is Cvg= €,,I, cos@= R J Z . We also have

cos@,where I,,

= &ms/Z and

AC Circuits

Combining these results we obtain

35-21

â&#x201A;Ź2R W 2

payg = m2R2+ L2(w2-mi)' (b) Energy dissipation will be a maximum, when dP,,/dw = 0. Taking the derivative, devg -dm

m2R2+ L2(w2-@:)I

02[2wR'

+ 2L'(W? - W 3 ( 2 W ) ]

* m2R2+ L2(w2-mi)' = U 2 R 2+ 2 L 2 W 2 ( 0 2-mi) + w4 = w,"a w = W, 35.69. Visualize: to Figure CP35.69.

Voltage is the same for circuit elements in parallel, so start with the E, phasor. Please refer I

t EO R

L I

Solve: (a) We see from the above phasor diagram that

Since I = E , / Z , the current is

(b)As w

+ 0 rads, WL + 0 R, so I + -.

That is, the inductor becomes a short for &o. On the other hand, as w -+

-, I + 00 because now the capacitor becomes a short for &,. (c) To find the frequency for which I is minimum, we set dI/dw = 0 . We get

The resonance frequency is the same as a series RLC circuit. (d) We know the current is a minimum at w = w,and diverges as look more or less as sketched above.

w +0

or w + -. Thus the current graph must

35-22

Chapter 35

35.70. Model: The capacitor current leads the voltage by 90". The inductor current lags the voltage by 90". Vosualize: (a) The capacitor and inductor are in parallel. Parallel circuit elements have the same voltage, so V, = Vc = VLc. Because this is a shared voltage, we've started the phasor diagram by drawing the V, phasor. Then the capacitor phasor I , leads by 90" and the inductor phasor ZL lags by 90".

I .=

The capacitor and inductor currents add to give the phasor ZLc = ZL - IC. The resistor is in series with the capacitorhnductor combination, and series elements share the same current, hence IR = ZLc. For a resistor, the voltage V, is in phase with the current I,, and this has been shown on the diagram. Finally, the voltages VR and V, add as vectors to give the emf voltage &., We see from the right triangle that &; = v,: + v$ For the resistor, V, = I,R = ZR. The capacitorhnductor combination requires a little more care. The peak inductor current and peak capacitor current are

Thus

But I , = I , = I , thus our expression for &, becomes

(b) As w + 0 rads, X,

+ 0 R and Xc + -.

This means

(+-+]+-

($4 -2

Thus, I = &,/R. That is, the inductor L becomes a short and E, drives only the resistor R. As w + 00, XL +

+ 0 R . This means

Thus. I = &o / R . That is. the capacitor C is the short and Eo drives only the resistor R. (c) At resonance I , = I , . This means X, = Xc or W , = Therefore,

Thus, I = 0 A.

and