Trying to apply superposition method to circuit, not working

[TheKShaugh]

Homework Statement

A double loop circuit is shown in the figure below.[/B]

The Emf

x = 6.00 V and the Emf

w = 19.00 V. Both are shown on the diagram. Calculate the power dissipated in the 6 Ω resistor located on the extreme right in the circuit.

Homework Equations

V=IR, P=VI

The Attempt at a Solution

I tried using superposition, where I replaced one battery with a short and solved for the voltage and current through the 6 ohm resistor, and the did the same by replacing the other battery with a short. The total current and voltage is the sum of the individual values. I plugged those final values into P=VI and got the wrong answer. Should this method work and I just did the work wrong or is there something I'm not aware of?

 

[phinds]

Sounds like you did it right but without your posting your work it's impossible for us to see where you might have gone wrong.

 

[rude man]

I tried using superposition, where I replaced one battery with a short and solved for the voltage and current through the 6 ohm resistor, and then did the same by replacing the other battery with a short. The total current and voltage is the sum of the individual values. I plugged those final values into P=VI and got the wrong answer. Should this method work and I just did the work wrong or is there something I'm not aware of?

You cannot add power by superposition. You can compute current I by superposition:
I = I1 + I2.
Then P = I^2 R.
Or you could have computed V1 and V2 by superposition, then P = VI = (V1 + V2)(V1 + V2)/R.

 

[phinds]

You cannot add power by superposition. You can compute current I by superposition:
I = I1 + I2.
Then P = I^2 R.
Or you could have computed V1 and V2 by superposition, then P = VI = (V1 + V2)(V1 + V2)/R.

DOH ! I assumed that's what he was doing but on rereading I see I just glossed over what he really said. Good catch.