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RC Circuit - Charge on Capacitor

  • Thread starter calvert11
  • Start date
1. Homework Statement
In the figure below the battery has an emf of 20 V and an internal resistance of 1.
Assume there is a steady current flowing in the circuit.
http://img48.imageshack.us/img48/9211/44724253.gif [Broken]
Find the charge on the 4 μF capacitor.
Answer in units of μC.

2. Homework Equations

Q = CE

3. The Attempt at a Solution
No time is given, so I can't use the equation for charge as a function of t.

From the figure, Resistors 1 and 5 seem to be in series. The circuit can't be simplified further.

But the charge on the capacitor can't be Q=CE since a current is still flowing. Can I have some advice on how to get started?

I tried using Kirchhoff, but I didn't get anywhere. I think I have too many unknowns, not enough equations (am I wrong?)

I'd be very grateful for any help.
 
Last edited by a moderator:

LowlyPion

Homework Helper
3,079
4
It's true there is current flowing in the system, but there is still a voltage across the capacitor.

Q = V*C

Determine then the voltage across the capacitor and the charge on the capacitor will be known.
 
It's true there is current flowing in the system, but there is still a voltage across the capacitor.

Q = V*C

Determine then the voltage across the capacitor and the charge on the capacitor will be known.
Okay, using I= V/R the current for equivalent resistor 6 is 3.333 and for resistor 8 it's 2.5.

Voltage at resistor 5 is 3.333*5=16.6665
Voltage at resistor 8 is 2.5*8 = 20
Voltage through the capacitor is 20 - 16.6665 = 3.3335, which is wrong.

What about my approach is incorrect?
 

LowlyPion

Homework Helper
3,079
4
Okay, using I= V/R the current for equivalent resistor 6 is 3.333 and for resistor 8 it's 2.5.

Voltage at resistor 5 is 3.333*5=16.6665
Voltage at resistor 8 is 2.5*8 = 20
Voltage through the capacitor is 20 - 16.6665 = 3.3335, which is wrong.

What about my approach is incorrect?
For 1 thing V = I*∑R = 20/14

That makes Vc = 8*20/14

Q = V*C = 4μf*160/14 = 640/14 μf
 
For 1 thing V = I*∑R = 20/14

That makes Vc = 8*20/14

Q = V*C = 4μf*160/14 = 640/14 μf
That's incredibly helpful. But to clarify, resistors 1, 5, and 8 are actually in series because no current flows through the wire with the capacitor?
 

LowlyPion

Homework Helper
3,079
4
That's incredibly helpful. But to clarify, resistors 1, 5, and 8 are actually in series because no current flows through the wire with the capacitor?
There was current when it charged of course. But basically with the capacitor all you are concerned with is its voltage. And that is determined by the Voltage Divider of the resistors in series.
 

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