Desperately need help with part B only of this problem. Cannot seem to get the c

Question

Desperately need help with part B only of this problem. Cannot seem to get the calculation correctly for some reason.

A light bulb has a resistance of 200 O. It is connected to a standard wall socket (120 V rms, 60.0 Hz).

(a) Determine the current in the bulb.
Answer: 0.6A

****(b) Determine the current in the bulb after a 14.5-µF capacitor is added in series in the circuit.*****


?????

(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?

0.485 H

Explanation / Answer

impedance due to capacitor = 1/C = 1/(2*60*14.5*10^-6) = 182.9

impedance due to resistor and capacitor = ((182.9)^2+200^2)^1/2 = 271.02

current = 120/271.02 = .442A

f = 1/2LC

usng this we get L = .0485H

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