# Find I through one of three resistors in a 3-battery circuit

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1. Apr 5, 2016

### kamhogo

1. The problem statement, all variables and given/known data

Sorry for the quality of the image. This is the best I can come up with with my cellphone.

What is the current through the 10 ohms resistor?
Is the current from left to right or right to left?

2. Relevant equations
Kirchhoff's loop law
Kirchhoff's junction law

3. The attempt at a solution

1) I1 = I2 + I3 ==> I3 = I1 -I2

2) -5(I1) + 12V - 10(I2) + 3V = 0
15V = 10(I2) + 5(I1)
I1 = (15V - 10(I2))/5

3) -5(I3) + 9V + 3V + 10(I2) = 0
-5(I1-I2) + 12V + 10 (I2) = 0
12V = -15 (I2) + 5 (I1)
= -15 (I2) + 15V - 10 (I2)
-3V = -25 (I2) ===> I2 = 0.12 A, from right to left.

This answer is incorrect. The correct answer is 0.36 A. Can someone please give me a hint as to what I'm doing wrong ?

2. Apr 5, 2016

### Simon Bridge

Check the signs of the voltages in (3).

3. Apr 5, 2016

### kamhogo

3) -5 (I3) + 9V - 3V + 10(I2) = 0
-5 ( I1-I2 ) + 6V + 10 (I2) = 0
-5 (I1) + 5 (I2) + 6V + 10(I2) = 0
6V = 5 (I1) - 15 (I2)
6V = 15V - 25 (I2)
-9V/-25 ohms = I2 = 0.36A

I got to the right answer by trial and error. I don't understand why the 3V has to be negative when the current ( I2) is flowing through it from its negative terminal to its positive one. ...Can someone please explain?

4. Apr 5, 2016

### Staff: Mentor

The potential difference across an ideal voltage supply does not depend upon the current or the direction of the current flowing through it. When you do your "KVL walk" around the loop you add or subtract the supply voltage strictly depending upon the direction that you pass through the fixed potential difference.

5. Apr 5, 2016

### kamhogo

Ok. I get it. Thanks!

6. Apr 5, 2016

### Simon Bridge

What I usually do is draw potential difference arrows next to the components ... the arrow points from the low potential end to the higher potential end: so it points from negative to positive by a battery and the opposite direction to the (proposed) current by a resistor.
When I traverse the loop, I add potential differences where I go in the direction of the arrow and subtract the ones where I go against the arrow.

7. Apr 5, 2016

### kamhogo

Thanks for the tip!