Alternating current circuits

[arod2812]

Homework Statement

A light bulb has a resistance of 212 . It is connected to a standard wall socket (120 V, 60.0 Hz). (a) Determine the current in the bulb. (b) Determine the current in the bulb after a 13.2-μF capacitor is added in series in the circuit.

Homework Equations

R, V, f are given. I don't know how to start incorporate these figures without a value for Z. Am I forgetting a another important equation?

The Attempt at a Solution

I used the equations:
X(sub L)=2pifL
Irms=Vrms/Z

 

[cepheid]

The Attempt at a Solution

I used the equations:
X(sub L)=2pifL

Why? There is no inductor in the circuit!

Irms=Vrms/Z

Okay, good. Z is the impedance. What is the impedance of the circuit with just the resistor? What about after adding a capacitor?

 

[Moetasim]

Z=sqrt(R^2 + X(sub L)^2)

(a) To determine the current in the bulb without the capacitor, we can use the equation Irms=Vrms/Z, where Vrms is the root mean square value of the voltage (120 V) and Z is the total impedance of the circuit. Since the bulb has a resistance of 212 Ω, we can calculate the impedance as Z=sqrt(R^2 + X(sub L)^2), where X(sub L) represents the inductive reactance of the circuit, given by X(sub L)=2pifL. The frequency is given as 60.0 Hz and the inductance of the circuit can be assumed to be negligible, so X(sub L)=0. Therefore, the total impedance is simply equal to the resistance, Z=212 Ω. Plugging in these values, we get Irms=120 V/212 Ω=0.566 A. Therefore, the current in the bulb without the capacitor is 0.566 A.

(b) To determine the current in the bulb after the capacitor is added in series, we can use the same equation, Irms=Vrms/Z, but this time the impedance will be different. Since the capacitor is now in series, the total impedance of the circuit will be the sum of the resistance and the capacitive reactance, given by X(sub C)=1/(2pifC). The frequency is still 60.0 Hz, but now we have a capacitance of 13.2 μF, which can be converted to Farads as 13.2x10^-6 F. Plugging in these values, we get X(sub C)=1/(2pifC)=1/(2x3.14x60.0x13.2x10^-6)=201.5 Ω. Therefore, the total impedance is now Z=212 Ω + 201.5 Ω=413.5 Ω. Plugging this into the equation, we get Irms=120 V/413.5 Ω=0.290 A. Therefore, the current in the bulb with the capacitor added is 0.290 A. We can see that the current has decreased due to the addition of the capacitor, which acts as a reactive element in the circuit and decreases the overall impedance.