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