What is the charge on the capacitor?

  • #1

Homework Statement



https://www.physicsforums.com/attachment.php?attachmentid=20071&stc=1&d=1250025514

The switch in the figure has been closed for a very long time.
https://www.physicsforums.com/attachment.php?attachmentid=20071&stc=1&d=1250025514

(a) What is the charge on the capacitor?

(b) The switch is opened at t = 0 s. At what time has the charge on the capacitor decreased to 10% of its initial value?


Homework Equations



E-iR1-iR2=0
emf (delta)Vi=0
Q=Qe^-t/RC (possibly)

The Attempt at a Solution


So, I know that the circuit is initially closed, and I don't need current to solve. Also that I probably need to combine resistors because they are in series. Yet, I'm a little confused as how to set up the problem.

Any type of help or insight would be deeply appreciated.
 
Last edited:

Answers and Replies

  • #2
tms
644
17


You'll have to at least describe the circuit in words, if you can't post a graphic of it.
 
  • #3
125
0


If we could see the picture it might help. If you can not load describe the circuit and value of the components. This would help a lot.
 
  • #5


It looks something like this:


____WWW______________
I..........60 OHM.....I................I
I...........................I.........…
_ 100V.................W...........W
-.............40 OHM W............W 10 OHM
I............................I........…
I............................I........… I_
I............................I........… 2 micro F
I............................I........…
I________________I_________I
 
  • #6
412
4


Your picture is still not clear, I couldn't find where is the switch but general suggestions:
1) Convert your circuit to thevenian with capacitor at output (with one voltage source and one thevenian resistor)
2) 1/RC it T(time constant)
3) Use x(inf) - [x(inf) - x(0)] exp (-t/T) formula for finding voltage across capacitor as a function of t.
4) For steady state (when long time), make capacitor open and solve the circuit
 

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