# Power Dissipation (more than one battery)

1. Apr 16, 2010

### ReMa

1. The problem statement, all variables and given/known data

For the circuit shown, find the power dissipated by the 10.0 ohm resistor.

(as attached)

2. Relevant equations

P=IV=I2R=V2R
V=IR

3. The attempt at a solution

I really don't know where to start with this. I'm basically tied up in the fact that there are three different sources of voltage.

Am I to add the series/parallel resistances up for the whole circuit? I think I have to find current I in order to solve this but i'm really confused.

#### Attached Files:

• ###### powerdis.bmp
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2. Apr 16, 2010

### willem2

None of the resistances or voltage sources is in series or parallel as the circuit is now.

However the positive sides of the 10 ohm voltage sources are at the same potential, so you can connect them with a wire without affecting any of the currents or voltages in the circuit.
two of the resistors and two of the voltage sources are now parallel.

(If you have two equal voltage sources in parallel you can replace them with a single source of the same voltage. If you have two unequal voltage sources in parallel you get a short circuit)

3. Apr 16, 2010

### emanuel_hr

To solve this kind of problems, it is customary to choose one node as a reference(0V). After that you can calculate potentials with respect to that node and consequently power dissipations in resistors.
My sugestion is to choose the node where the 3 voltage sources meet as your 0V referece.