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Drakon25th
May13-04, 08:18 PM
Ok, the question says: "Find the magnitudes and directions of the current through the 22 ohm and 15 ohm resistors (the internal resistance of the batteries is negligible)"

My question is how do i determine this if there's a second emf in the problem?

here's a picture of the problem:
picture (http://mysite.verizon.net/vze3ss2y/sitebuildercontent/sitebuilderpictures/emf.jpg)

i tried using Kirchoff's loop law at first, first determining these two formulas:
0=E1-I1R1-I2R2
=9.0 - 22I1 - 15I2
I1 = .41 - .68I2

and

0=E2 - I2R2
= 6-15I2
I2 =.40

The problem is the second equation doesn't incorporate I3, so i cant use I3 = I2 + I1, so I'm stuck here.
Then I thought about using Ohm's law, but I am not sure how do i use E1 in terms of the parallel circuit to determine the total current going through the whole circuit. Please help me someone :smile:

Drakon25th
May13-04, 08:18 PM
by the way, the directions for I1, I2, and I3 in the picture are my predictions on the current flow throughout the circuit

Drakon25th
May13-04, 08:43 PM
bump, please help me =(

Janitor
May13-04, 09:04 PM
I'm no electrical engineer, but...

It seems to me that given your choices for current flow (and I am presuming you are using conventional current, which imagines that a positive charge is what is flowing), your first equation has a bogus sign and should be:

0=E1-I1R1+I2R2.

Does it help to write another equation based on the third path of flow? That is the one with the two batteries and R1 in it:

E1+E2=I1R1.

That would allow you to immediately solve for I1. Then You could put that value in my corrected version of your first equation and solve for I2.

Drakon25th
May13-04, 09:08 PM
so there would be no I3?

Janitor
May13-04, 09:09 PM
If my analysis is correct, then it gives you I1 and I2, and as you said, you can then get I3 by adding those two: I3 = I2 + I1.

Drakon25th
May13-04, 09:13 PM
alright, thank you so much for the help :)