How Does Adding a Capacitor Affect Current in an AC Circuit?

AI Thread Summary
The discussion revolves around calculating the current in a light bulb connected to a 120 V, 60 Hz AC circuit, initially with a resistance of 212 ohms. Participants are seeking to understand how to incorporate a capacitor, specifically a 13.2-μF capacitor added in series, into their calculations. The key equations mentioned include the root mean square current (Irms = Vrms/Z) and the need to determine the impedance (Z) of the circuit. There is confusion regarding the use of inductive reactance, as no inductor is present in the circuit. Ultimately, the focus is on accurately calculating the current before and after the capacitor is added.
arod2812
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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
 
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arod2812 said:

The Attempt at a Solution


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

Why? There is no inductor in the circuit!
arod2812 said:
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?
 

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