Capacitor Question Homework: Finding the Potential Difference

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The discussion revolves around a homework problem involving capacitors and circuit analysis. The equivalent resistance was calculated as 68 Ohms, leading to a current split where the capacitor experiences a potential difference of 0 V after the switch has been closed for a long time. This indicates that the capacitor behaves like an open circuit, resulting in no current flowing through it or the 10 Ohm resistor. The voltage across the 40 Ohm resistor is equal to that across the capacitor, confirming the static state of the circuit. The role of the 10 Ohm resistor is questioned, suggesting it may influence the charge rate when the switch is closed.
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Homework Statement


2mhwndg.jpg




Homework Equations


V = IR
CV = Q



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.
 
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"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.
 
Oh...so then no current also flows through the 10 Ohm resistor?
 
You've got it. well done
 
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
 
Snazzy said:
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.
 
What's the point of that 10 Ohm resistor then?
 
I dunno. It would effect the charge rate when the switch is closed. Ask your prof.
 

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