# RLC Circuit - Voltage drop across Inductor

1. Apr 29, 2009

### calvert11

1. The problem statement, all variables and given/known data

A 10 μF capacitor and a 25 H inductor are connected in series with a 60 Hz source whose rms output is 112 V.

Find the voltage drop across the inductor. Answer in units of V.
Note: Sigfigs do not matter

2. Relevant equations

I = V/Z where Z is the impedence

XL = omega*L
XC = 1/omega*C

3. The attempt at a solution

I assumed the closed circuit was connected thus: battery to the capacitor, the capacitor to the inductor, the inductor to the battery.

I found the current using equation I = V/Z (the answer was correct)

I found the voltage drop across the capacitor by the equation V - I*XL = $$\Delta$$V (the answer was correct)

For the voltage drop across the inductor I tried I*XC - V. The answer is incorrect.

Could anyone tell me what I'm doing wrong?

Last edited: Apr 29, 2009
2. Apr 29, 2009

### davieddy

The voltage across the inductor leads the current by pi/2.
The voltage across the capacitor lags the current by pi/2.

So the voltages are in antiphase, and it is the difference between them
(not the sum) which is equal to the source voltage.

But having found the current, you can simply apply pd = I*XL to get the
voltage across the inductor.

Last edited: Apr 29, 2009
3. Apr 29, 2009

### calvert11

Thank you, this explains a lot. My answer would have been "pd = I*XL" but I thought it didn't make sense since it was greater than the source voltage.