# Circuit Related Q

1. Jul 24, 2006

### MD2000

Determine the power dissipated in the 6.0 resistor in the circuit shown in the drawing. (R1 = 4.0 , R2 = 6.0 and V1 = 15 V.)

Am I supposed to be using P = V^2/R..is the power flowing through each resistor gonna be diff?

Anyone wanna help me out where to start?

2. Jul 24, 2006

You can solve this many different ways. First note that the current through the 2ohm and the 1ohm resistor is the same, and that the current through R2 and the 1ohm resistor is the same.

I would first solve for the currents, and then you know that:

$$P=IV$$

$$V=IR$$

Thus,

$$P=I(IR)=I^2R$$

Do you know KVL, KCL?

3. Jul 25, 2006

### sdekivit

4. Jul 25, 2006

### MD2000

How exactly can I get the currents for each individual peice?

I know that the total current is gonna be V = IR..which equals 2.46..

But how can I solve for individual currents? Since you don't have V..what can you use to get the I?

Last edited: Jul 25, 2006
5. Jul 26, 2006

### Hootenanny

Staff Emeritus
Think Kirchoff's laws, you can use these the calculate the voltage or current at each point in the circuit.

6. Sep 8, 2006

### physics_girl

I have a similar question with multiple batteries and resistors in one big circuit. I have been given two labelled junctions but within one smaller loop, there are two batteries and one resistor...
I don't know where to begin, any advice???