# Homework Help: Circuit problem

1. Feb 7, 2005

### Silimay

I have a circuit problem I'm having trouble with. I've attached an image of it.

$$Emf_1 = 3 V; Emf_2 = 1 V; R_1 = 5 \Omega; R_2 = 2 \Omega; R_3 = 4 \Omega$$

Both batteries are ideal. a.) What is the rate at which energy is dissipated in $$R_1$$?

Since $$P = i^2R$$, I figured that I should find the current. I used the loop rule (I'm not sure if I should have done this or not). I started at a point between $$emf_1$$ and $$R_3$$. I pretended the current was pointed downward at $$emf_1$$ and upward through $$R_1$$ and $$R_2$$. I called those last two currents $$i_1$$ and $$i_2$$, respectively. $$i_{total} = i_1 + i_2$$.

For the inside (left) loop:

$$R_3i_{total}+R_1i_1+emf_1 = 0$$

For the outside loop:

$$R_3i_{total} + R_2i_2-emf_2+emf_1 = 0$$

$$+R_3(i_1+i_2)+R_1i_1 = -emf_1$$

$$-R_3(i_1+i_2)-R_2i_2+emf_2=-emf_1$$

$$(5\Omega)i_1-(2\Omega)i_2 = -7 V$$

$$-(4\Omega)(i_1+i_2)-(2\Omega)i_2 = -3 V$$

$$-(4\Omega)i_1-(6\Omega)i_2 = -3 V$$

I solved the system of equations and ended up with $$i_1 = 0.95 A$$. This isn't the correct answer, because $$i^2R$$ ends not equalling 0.346 W, which is the right answer.

I have the feeling I did something really wrong. Did I make it way too complicated? I have a tendancy to overdo things

~Silimay~

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2. Feb 7, 2005

### jamesrc

Hi,

Two things I can see:

- With the way you defined things and the way you are going around your loops, the signs on your voltages should be switched (e.g. first loop eq should be R3itotal1+R1ii = 0).

- I think you made some algebra mistakes after that.

Try again and I think you'll get it.

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