# Homework Help: Circuit with 2 sources of emf

1. Nov 18, 2004

### joshanders_84

I have problem with a circuit. It has two sources of emf, and I'm not sure how to calculate the magnitued of the current in the circuit due to this. Here's what the circuit looks like (R's are resistors, *'s are just for spacing):

|----------+|1|------|
|*************** |
R = 5 Ohms********R = 9 Ohms
|*************** |
|----------+|2|------|

EMF source 1 has resistance: 1.6 Ohm, and is 16 V
EMF source 2 has resistance: 1.4 Ohm, and is 8 V

I tried calculating them both and subtracting one from the other, but I don't know. I know the top one is stronger, so the current is traveling in a counter clockwise manner, but don't know the technique when there are two emf sources for solving the magnitude of the current. Thanks!
Josh

Last edited: Nov 18, 2004
2. Nov 19, 2004

### vsage

Treat the EMF sources and their resistances as separate components and apply the loop rule from any point on the circuit (I suggest using one of the EMF sources). Remember the current is constant throughout the system:

$$16V - IR_1 - IR_2 - 8V - IR_3 - IR_4 = 0$$

I applied the law counterclockwise starting from EMF 1 but like I said you could start from anywhere. Solving for I you obtain the familiar

$$I = \frac{\varepsilon_1 - \varepsilon_2}{R_1+R_2+R_3+R_4}$$ which looks a lot like

$$I = \frac {\sum \varepsilon}{\sum R}$$

Hope this helps. Sorry but I don't know the LaTeX for that pretty little E my physics book uses so I figured lowercase epsilon suffices :)

Last edited by a moderator: Nov 19, 2004
3. Nov 19, 2004

### Nylex

It's still an epsilon and you use \varepsilon.

4. Nov 19, 2004

### cepheid

Staff Emeritus
Oh I see...the notation for the EMF's of the sources: try the \mathcal function--it gives you uppercase scripted characters, should you need them:

$$\mathcal{E}$$

Sometimes for source voltages we just write $V_s$ instead. But since these are not ideal sources, it's good to distinguish between the EMF, which is defined as the potential difference between the two source terminals when no load is connected, vs. the actual voltage across the source when in this series circuit. If you already knew all of this...sorry to bore you to tears.

I like these scripted letters...hmm...let's see...Laplace Transform:
$$\mathcal{L} \{f(t)\}$$

it's cool...

5. Nov 19, 2004

### vsage

I see. Thanks for the info on that. Post has been edited.