A light bulb has a resistance of 250 . It is connected to a standard wall socket

Question

A light bulb has a resistance of 250 . It is connected to a standard wall socket (120 V rms, 60.0 Hz). (a) Determine the rms current in the bulb. Correct: Your answer is correct. A (b) Determine the rms current in the bulb after a 19.5 µF capacitor is added in series in the circuit. Incorrect: Your answer is incorrect. A (c) It is possible to return the current in the bulb to the value calculated in part (a) by adding an inductor in series with the bulb and the capacitor. What is the value of the inductance of this inductor?

Explanation / Answer

a)

Rms current

Irms =Vrms/R =120/250

Irms=0.48 A

b)

Capacitive reactance

Xc=1/2pi*f*C =1/2pi*60*(19.5*10^-6)

Xc=136.03 ohms

impedance

Z=sqrt[R^2+Xc^2] =sqrt[250^2+136.03^2]

Z=284.6 ohms

Rms current

Irms=Vrms/Z =120/284.6

Irms=0.42 A

c)

at resonant frequency it is possible

f=1/2pi*sqrt(LC)

=>L=1/4pi^2*f^2*C =1/4pi^2*60^2*(19.5*10^-6)

L=0.36 H

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