# Current through resistor parallel to two different batteries

1. Sep 9, 2014

### Ellie

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

I'm sorry that I have no pic, but please bear with my description.

The circuit diagram is made up of 3 horizontal lines. On the top line, there's battery A, (-) on the left, (+) on the right, and an arrow going to the right (>) to show direction of current.

On the middle line, there's the 2 Ω resistor.

And on the bottom line, there's battery B, (+) on left, and (-) on right.

Battery A has an e.m.f, of 2.0 V and an internal resistance of 1 Ω. For battery B, the values are 1.0 V and 2 Ω. A and B are connected to a 2 Ω resistor. Using Kirchoff's Law, calculate current through the resistor.

2. Relevant equations

Both of Kirchoff's Law.
I1 = I2 + I3
E = IR1 + IR2

3. The attempt at a solution

I tried taking the top and middle, and treated it as one circuit, and did the same with the middle and bottom. Then I applied Kirchoff's second law. In this way, I managed to get two equations with three different variables.

2 * I3 + I1 = 2
- 2 * I3 + 2 * I2 = 1

Where I1 is the current leaving battery A, I3 is the current entering the middle section after entering a junction, and I2 is the other current going toward the bottom section.

Then by using I = I1 + I2 + I3, as well as the two equations, I substituted the values around, until I managed to make the equation in terms of I3.

The value I got is 0.375 A.

The answer is supposed to be 0.42 A.

It would be great if someone can point out where I did wrong.

2. Sep 9, 2014

### Tanya Sharma

Welcome to PF !

As per the circuit description you have given , 0.375 A is the correct answer . So I think you have got it right .

Either the answer key is wrong or the original question description is somewhat different from what you have given.

3. Sep 9, 2014

### Ellie

Thank you for clarifying. I'll see if I've misunderstood somewhere.

4. Sep 10, 2014

### rude man

Tanya has it right.