Power in 6.0 Ohm Resistor - Circuit Related Q

[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 going to be diff?

Anyone want to help me out where to start?

 

[FrogPad]

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?

 

[sdekivit]

there's exactly the same problem, only with different numbers here. You should check that out first:

https://www.physicsforums.com/showthread.php?t=126987

 

[MD2000]

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

I know that the total current is going to 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?

 

[Hootenanny]

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

I know that the total current is going to 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?

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

 

[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?