AP PHY Circuit Help: Solve 5 Questions

[RyanCouillard]

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.

 

[kamerling]

A. I don't think the circuit HAS a capacitance. I suppose you are meant to compute the capacitance of the 2 capacitors in series. I don't understand your answer at all, but the right number isn't in there.

B I suppose you're meant to calculate the current after the circuit is in equilibrium. In that case there is no more current going through the capacitors, which should make it easy to compute the current through the resistances.

C is easy if you solved B

E Once the circuit is in equilibrium, there's no current going through the capacitors, so what will happen if you cut the wire through them?

 

[Moetasim]

I would like to provide some guidance and clarification on the questions and calculations for this circuit.

A) To calculate the total capacitance of the circuit, we need to use the formula Ceq = C1 + C2 + C3 + ... where Ceq is the equivalent capacitance and C1, C2, C3 are the individual capacitances in the circuit. In this case, we have two capacitors in series (12 uF and 24 uF) and one in parallel (36 uF). The equivalent capacitance can be calculated as follows:

1/Ceq = 1/12 uF + 1/24 uF + 1/36 uF
1/Ceq = 1/6 uF
Ceq = 6 uF

B) To calculate the current in the 10 ohm resistor, we can use Ohm's Law (V = IR) and Kirchhoff's Current Law (the sum of currents entering a junction equals the sum of currents leaving a junction). First, we need to determine the total resistance in the circuit. The 20 ohm and 10 ohm resistors are in series, so their total resistance is 30 ohms. The equivalent resistance of the parallel combination of the two capacitors is 8 ohms (1/R = 1/12 + 1/24, R = 8 ohms). Therefore, the total resistance in the circuit is 38 ohms. Using Ohm's Law, we can calculate the current as follows:

I = V/R = 6V/38 ohms = 0.158 A

C) The voltage between A and B can be calculated using Kirchhoff's Voltage Law (the sum of voltages around a closed loop equals zero). We can start at point A and move clockwise around the loop. We have a 6V battery, a 20 ohm resistor, and a series combination of two capacitors (12 uF and 24 uF) with a total capacitance of 6 uF. Using the formula for capacitors in series (1/Ceq = 1/C1 + 1/C2), we can calculate the equivalent capacitance as 6 uF. Therefore, the voltage between A and B is:

6V - (20 ohms)(0.158 A) - (6 uF)(0.158 A)/6 uF =