Charge on capacitor (RC circuit)

  • #1

Homework Statement


1) Determine the current in each resistor and charge on the capacitor after a long time immediately after the switch is closed
2) Determine the current in each resistor and charge on the capacitor after a long time

After the switch has been closed for a long time it is re-opened.

3) Determine the current in each resistor and the charge on the capacitor immediately after the switch has been re-opened.
4) Determine the time constant of the capacitor discharge.

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Homework Equations


V = I Req
Q = CV


The Attempt at a Solution


I got #1 and got I6 = 1A, I8 = 1.5A, and I12 = .5A
I'm not sure what to do about #2 though. What is the voltage across the capacitor? I thought it would be 6V, because that is the voltage across the parallel series. However, there is a voltage drop after the 1A current runs across the 6ohm resistor so the V across the capacitor is 0.

For #3 and 4, does the 8ohm resistor come into play or no?
Thanks for the help guys.
 
Last edited:

Answers and Replies

  • #2
Is the V across the capacitor just 6V? If so I think I'm good. #2 I through the 6ohm resistor is 0, then I through the other two resistors is 18V/20ohms right?
#3 I = 6V/18ohms Q= 28 E-6 * 6
#4 (18 * 28E-6) = RC
 
  • #3
rl.bhat
Homework Helper
4,433
7
Voltage which charges the capacitor is the voltage across 12 ohm resistance. Calculate that voltage.
When the capacitor is completely charged, the same voltage will be the voltage across it.
 
  • #4
oh is this Kirchoff's rule?
 
  • #5
rl.bhat
Homework Helper
4,433
7
No. There is no need to apply Kirchhoff rule.
 

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