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Capacitor question

  • Thread starter Snazzy
  • Start date
458
0
1. Homework Statement
2mhwndg.jpg




2. Homework Equations
V = IR
CV = Q



3. The Attempt at a Solution
I found the equivalent resistance to be 68 Ohms, then I found the current, then I found that the current splits 4/5 of its total across the branch with the capacitor, but if I apply Kirchhoff's loop law, I just get that the capacitor has a potential difference of 0 V.
 

Answers and Replies

4,222
1
"The switch has been closed for a long time."

I don't know how this should be apparent an introductory level but...

With the switch closed for a while, everything is static, the currents are done changing because the capacitor has as much charged as it's going to get. So...

No current flows through the capacitor. It's, like, an open circuit.
 
458
0
Oh...so then no current also flows through the 10 Ohm resistor?
 
4,222
1
You've got it. well done
 
458
0
Okay, so if the 2nd branch doesn't affect the circuit, the current is 100 / (60+40) = 1A, and so the voltage difference across the 40 Ohm resistor is equal to the voltage difference across the capacitor? :S
 
4,222
1
Okay, so if the 2nd branch doesn't affect the circuit, the current is 100 / (60+40) = 1A, and so the voltage difference across the 40 Ohm resistor is equal to the voltage difference across the capacitor? :S
That's correct.
 
458
0
What's the point of that 10 Ohm resistor then?
 
4,222
1
I dunno. It would effect the charge rate when the switch is closed. Ask your prof.
 

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