In the figure, the ideal batteries have emfs 1 = 17.0 V and 2 = 0.5001 V, and the resistances are each 1.22 Ω.
I tried to include a link but I am unable.
As best as I can describe: There is one main loop with a wire connecting the bottom and the op piece. The two emfs are on the left and right side BOTH pointing towards the top (positive is both towards the top of the circuit/each other). R_1 is on the left side top and R_2 is on the right side top. R_3 is in the middle piece of connecting wire.
I hope that explains it well enough.
a) What is the current in resistor 2?
b) What is the current in resistor 3?
ƩΔV=0 for a complete circuit
Ʃi in a junction must be equal. This is to conserve charge.
The Attempt at a Solution
I am getting close but the double emfs are confusing me. I have another problem similar to this that I am also not able to do. I have the test Monday and I am certain there will be a double emf circuit problem because our professor is just that sort of guy. He also never covered this in class.
i_2 goes out of the positive terminal of ε_2 into R_2, i_1 goes out of the positive terminal of ε_1 into R_1, and i_3 goes straight down.
Using the junction rule I get:
i_3=i_1 + i_2
I tried using the loop rule for the right loop going in a ccw direction assuming the current is going in the same direction -ccw.
I get -i_1*(R_2)-i_3*(R_3)+ε_2=0
left loop: -i_1*(R_1)-i_3*(R_3)+ε_1=0
I then subbed the three equations into each other using wolfram alpha to check my work before I did it by hand and they are wrong. I am not sure what I am doing incorrect and I tried to follow an example from the book but I could not figure out how to apply that material to this case.
Thanks for the help. I will post more later as I attempt it for the third time today.