Homework ## Chapter

22: Alternating Current

Ch 22: 7, 11, 21, 31, 37, 46, 53

22.7 An electric light, connected to an ordinary 120 V AC wall outlet, uses 60 W of power. What are

a) the rms current through the light,

b) the resistance of the light, and

c) the maximum and minimum instantaneous values of the power and of the current?P = IV 60 W = (I) (120 V)

I_{rms}= 0.5 A, or

i = I(t) = I

_{max}cos 2 f t = I_{max}cos 120 tI

_{rms}= 0.5 AI

_{max}= (1.414) (0.50 A)

I_{max}= 0.707 A

I_{min}= 0Likewise,

V = V _{rms}= 120 VV

_{max}= (1.414) (120 V) = 170 VV

_{min}= 0P = I V

P

_{max}= I_{max}V_{max}= (0.707 A)(170 V)

P_{max}= 120 W

P_{min}= 0

22.11 A laboratory power supply provides AC electricity with a peak voltage of 27.0 volts. What is the rms voltage?V

_{rms}= 19.1 V

22.21 If 0.8 ampere flows through a capacitor when it is connected across a 120-volt, 60 Hz AC voltage source, what is the capacitance of the capacitor?

X

_{c}= 120 V / 0.8 A = 150

C = 1.77 x 10^{ - 5}F

C = 17.7 x 10^{ - 6}F = 17.7 F

22.31 If 0.20 ampere flows through a L = 200 mH (millihenry) inductor when connected across a 30-volt AC voltage source, what is the frequency of the source?

X

_{L}= 2 f L150 = 2 f (200 mH)

f = 150 / [2 (200 x 10

^{ - 3}H)]f = 119 Hz

22.37 [Note the change in the values of the capacitor and the frequency -- and the change in the order of the calculations or questions asked!]

Consider a series RCL circuit with R = 80 , C = 15 F, L = 250 mH connected to a 12 V, 120 Hz AC source.

a) Calculate the reactances X_{C}and X_{L}.

b) What is the total impedance?

c) What current flows through the circuit?

d) What are the three potential differences across the three elements (V_{R}, V_{C}, and V_{L})?

X_{C}= 88X

_{L}= 2 f LX

_{L}= 2 (120 s^{ - 1}) (0.250 H) = 188

X_{L}= 188

Z = 128

I = 0.094 A

The voltages we calculate across the circuit elements, like the 12 volts of the power supply itself, are all rms-values.

V _{R}= I R = (0.094 A)(80 ) = 7.52 VV

_{C}= I X_{C}= (0.094 A)(88 ) = 8.27 VV

_{L}= I X_{L}= (0.094 A)(188 ) = 17.67 VSome explanation or discussion of V

_{C}and, especially, V_{L}seems like it might be useful. The voltages across the capacitor and the inductor are out of phase with each other so the net voltage across the two of them isV _{net}= | V_{L}- V_{C}| = | 17.67 V - 8.27 V | = 9.4 V

22.46 [Note thechange in the valueof the inductance!]

An 80 resistor in connected in series with a 50 F capacitor and a 30 mH inductor across a 120 V, 60 Hz AC voltage source.

What is the impedance of the circuit?

How much current flows in the circuit?

What is the power factor (cos ) of the circuit?

How much power is used by the circuit?

X

_{C}= 53X

_{L}= 2 Ó f LX

_{L}= 2 (60 s^{ - 1})(0.030 H) = 11.3X

_{L}= 11.3Z = 90.2

I = 1.33 A

P = I

^{2}Z cos = I V coscos = power factor

cos = 0.887

P = (1.33 A) (120 V) (0.887) = 141.5 W

P = 141.5 W

22.53 What is the resonance frequency of a series RCL circuit with a 150resistor, a 25 F capacitor, and a 80 mH inductor? What current will flow in the circuit if it is connected to a 20 V supply at this resonance frequency? What current will flow if the frequency is increased or decreased by 20%?At resonance Z = minimum or Z = R, so

X _{C }= X_{L}f = 112 Hz

At this resonance frequency of 112 Hz, the total impedance is at its minimum,

Z _{min}= R = 150so the current through the circuit is

At resonance, the current is 0.133 A

I _{res}= 0.133 ANow

increasethe frequency by20%f _{+}= (1.20) f_{res}= (1.20) (112 Hz) = 134 HzX

_{C}= 47.5X

_{L}= 2 f L = 2 (134 Hz) (80 x 10_{ }^{- 3 }H)X

_{L}= 67.4Z = 151.3

Now,

decreasethe frequency by 20%,f _{-}= (0.80) f_{res}= (0.80) (112 Hz) = 89.6 HzX

_{C}= 71.5X

_{L}= 2 f L = 2 (89.6 Hz) (80 x 10^{ - 3}H)X

_{L}= 45.0Z = 152.3

And we see there is very little change in the current when the frequency is changed by

20%.It is interesting and fun to re-do this set of calculations for R = 15 (Remember, for the case we have just looked at, we had R = 150 ). This new circuit has a much

smallerresistance R.The resonance frequency remains the same, of course. The values of the reactances, X

_{L}and X_{C}, remain the same. Only R has changed. At resonance, the new resonance current is nowI

_{res}= 1.33 A

For the higher frequency, f

_{+}= 134 Hz, we still haveX _{C}= 47.5 and X_{L}= 67.4Z = 24.9

Now, decrease the frequency by 20% again, so that f

_{-}= 89.6 Hz, and we still haveX _{C}= 71.5 and X_{L}= 45.0Z = 30.4

Now -- with a smaller value of the resistance R -- a variation in the frequency of 20%

does, indeed, make a verylargedifference in the current.(c) Doug Davis, 2003; all rights reserved

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