# Finding charge of a capacitor in RC circuit

1. Mar 27, 2013

### Teptip

Thanks, solved!

Last edited: Mar 28, 2013
2. Mar 28, 2013

### ehild

Yes, after a short time that the circuit was connected the capacitor is charged and does not take current any more.

The capacitor is connected parallel to the 4 ohm resistor. What do you know about the voltage across parallel connected circuit elements?

ehild

3. Mar 28, 2013

### Teptip

Okay so I can calculate current I=V/R = 19/16 = 1.1875 A

V total = (V of Capacitor) + (V of 11Ω Resistor)

V of 11Ω Resistor = IR = (1.1875)(11) = 13.0625

19 = (V of Capacitor) + 13.0625
V of Capacitor = 5.9375

Then plugging into the capacitance formula:
Q = CV = (4µ)(5.9375) = 23.75µ

Does this work look good?

4. Mar 28, 2013

### ehild

The current is correct.

No, what do you mean of Vtotal? And you do not need it.

You know the voltage across the 11 Ω resistor. Look at the figure. Where are the plates of the capacitor connected to?

ehild

5. Mar 28, 2013

### Teptip

Is the voltage I want then just the voltage across the 4Ω resistor?

6. Mar 28, 2013

### ehild

Yes. And what current flows though it?

ehild

7. Mar 28, 2013

### Teptip

Ah is it the same current as throughout, since the capacitor is quickly charged?

(I = 1.1875 A so V = (1.1875)(4) and then this voltage is the same that would go through the capacitor?)

8. Mar 28, 2013

### ehild

Yes, they are connected to the same points, so the voltage must be the same across both. (The voltage does not "go through", it is across two points)

ehild