There is a diagram for this question, but I can't post it, so I just describe it. There are four resistors R1,R2,R3,R4. When the switch S is opened, current only flows through R1,R3 and R4. R3 and R4 are parallel to each other, but in series with R1. Now the switch is closed, what happens to the voltage across each resistor?
My solution is first to find out the current flows before and after the switch is closed through each resistor. I redrew the circuit into two circuits in series. First one is when switch is opened (R1,R34--I combined R3,R4 in parallel), the second one is when switch is closed (R1,R234--combined R2 with R34). Then I figured out that R234 < R34. Since V=IR. Now I am looking for the change in current before and after. My teacher told me current after the S is closed is bigger than the current when the S is opened? BUt is that because when you add more resistance to the circut, the current through the circuit will decrease? which in this case R234<R34, that's why I is more when the switch is closed?

Thanks.

host your drawing and post the address. (well post most of it anyway)
Perhaps this site will help you. http://imageshack.us/

You should be able to post links very soon anyway, your only a few posts away from that.

Based on how you described it you have R3 and R4 in parallel but both in series to R1. If the switch, when open, only allows current to flow through R1, R3 and R4 then that means the switch is in series with R2. Now reading further it appears that R2 (when brought into the circuit by closing the switch) is in parallel with R3 and R4 and all three are in series with R1.

The more resistors that are in parallel, the less total resistance they offer. The less resistance you have at a constant voltage, the more amperage will flow across the circuit.

Show with values.

V = 12
R1 = 3 ohms
R2 = 1 ohm
R3 = 2 ohms
R4 = 3 ohms

With the switch open (no current path to R2)...

First find the total R of R3 and R4. 1/2 + 1/3 = 3/6 + 2/6 = 5/6 -> 6/5 = 1.2 ohms. Add that to R1 to get 4.2 ohms

12/4.2 = 2.857 amps

Now with the switch closed (current path to R2)...

Again find the total R of R2, R3 and R4. 1/1 + 1/2 + 1/3 = 6/6 + 3/6 + 2/6 = 11/6 -> 6/11 = .545 now add R1 to get 3.545

12/3.545 = 3.385 amps

Hi, Precursor562, thank you. I think you proved my point that since R234<R34, that's why I is more when the switch is closed.