- #1

RyanCouillard

- 2

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This one has me stumped. Here is the circuit

http://img90.imageshack.us/img90/6204/circuitdn1.jpg

I will go through all the 5 questions.

A) Calculate the total capacitance of the circuit.

18 uF=.5CV^2, 36uF=CV^2, 36uF/36v=1uColoumb. Correct?

B) Calculate the current in the 10 ohm resistor.

This is where we're stuck, there's going to be current coming off the capacitors, but which was will it flow? They have more potential difference than the battery, and I'm not sure how to figure that part out.

We have - I(20 ohm + 10 ohm)=6v, which would be 6/30=I. But this is only one loop, and depending on where the current flows the 20 ohm resistor will have a different current.

C) Calculate the voltage between A & B. (Forgot to add it, it is at the junction of the capacitors on either side.

(20 ohm) (6/30A)=4v + 12v (capacitors in series) = 18v

D) Calculate the charge stored on one plate of the 6 uF capacitor.

Ue=.5QV = 12uF=QV, and 12uF/6v=2uC

E)Wire is cut at p (Point b/w two capacitors), will the voltage increase, decrease, or remain the same?

It would decrease, and it would be 4v after? (6/30A)(20Ohms)=4v.

Thanks in advance, I have no idea on B.

http://img90.imageshack.us/img90/6204/circuitdn1.jpg

I will go through all the 5 questions.

A) Calculate the total capacitance of the circuit.

18 uF=.5CV^2, 36uF=CV^2, 36uF/36v=1uColoumb. Correct?

B) Calculate the current in the 10 ohm resistor.

This is where we're stuck, there's going to be current coming off the capacitors, but which was will it flow? They have more potential difference than the battery, and I'm not sure how to figure that part out.

We have - I(20 ohm + 10 ohm)=6v, which would be 6/30=I. But this is only one loop, and depending on where the current flows the 20 ohm resistor will have a different current.

C) Calculate the voltage between A & B. (Forgot to add it, it is at the junction of the capacitors on either side.

(20 ohm) (6/30A)=4v + 12v (capacitors in series) = 18v

D) Calculate the charge stored on one plate of the 6 uF capacitor.

Ue=.5QV = 12uF=QV, and 12uF/6v=2uC

E)Wire is cut at p (Point b/w two capacitors), will the voltage increase, decrease, or remain the same?

It would decrease, and it would be 4v after? (6/30A)(20Ohms)=4v.

Thanks in advance, I have no idea on B.

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