Help with capacitors, resistors, and batteries...?

Question

Can you please help me with these? Book doesn't help much.

1) http://img172.imageshack.us/img172/9498/3068og0.jpg
What is the C_eq of the 6 capacitors?
What is the potential diff between A and B?
** tried adding up the middle set of capacitors (series then parallel) first then the outside ones (parallel). Wrong. Answer is supposed to be 3C/2. How is E=0?

2) http://img172.imageshack.us/img172/5199/3112jf5.jpg
What is potential difference at the 10 ohm resistor? at 20 ohm?
**Tried V=IR, I=2, then V_10=2*10=20. This is wrong. Don't know why.

3) http://img172.imageshack.us/my.php?image=3150nu2.jpg
What is the value of the resistance R?
What is power dissipated by R?
**Tried I=sum (batteries) /sum(R), then V_R=IR =6, this is wrong. Need this to find power.

4) http://img172.imageshack.us/img172/443/3168vf3.jpg
The current and potential difference at 3 ohm?
**Tried adding sides to make series. V=12, R=71, I=1. Somehow wrong.

Answer 1

Total of C is 3/2 because when you connect two C in series, you get (1/2)C, then you have 3 of these in parallel. In parallel, you add them. (1/2)C * 3 = (3/2)C.

The voltage is ZERO because at both circuit involving point A and B, voltage is equally divided. Therefore, the difference between A and B is ZERO (assuming this is not a circuit with battery - read on)

BUT! This is a trick question. Recognize that power source is DC, and a capacitor does not let DC power through. So whatever you do, it always be ZERO volts. If the power source was AC, the above calculation would be true. I think, this may be a misprint on your problem itself. Check with your teacher if it was really meant to be a DC circuit.

If you have 15 volts source and 10 ohm and 20 ohm in series, you know, the total resistance is 30 ohms. The current flowing through it is (I = E/R) 15/30 or 0.5amps. If you want to figure out the voltage drop at 10 ohms, use E=IR then e = 10 * 0.5 = 5 volts. At the 20 ohms resister, since the source is 15 volts, and 5 volts drop at the 10 ohm, the remainder is 10 volts.

For 3, you are missing something. R can be any value with the information given.

For 4, you will need to rewrite the circuit diagram. Then you will see, 4 ohm, 48 ohm, and 16 ohms are in parallel. All of them together, it makes 3 ohms. Since then, 3 ohms is in series with another 3 ohms, you have 6 ohms across the battery. This is easy. Voltage is equally divided, so across 3 ohms, you have 6 volts.

It seems you have parallel and seires computations for capacitors and registers all mixed up. Also, try rewriting the circuit diagram. The way the question is written is confusing. All of them will reduce down to simple seires parallel circuit.

Answer 2

battery i'd say :) good drawing!! x