I don't want the answer, just a nudge in the right direction. I know that current I=ε/R(eq). Why wouldn't I= (ε1+ε2)/(R1+R2+R3)?
Consider the circuit in the figure below with resistors R1 = 2,400 Ω, R2 = 4,400 Ω, and R3 = 1,900 Ω, and batteries with emfs 1 = 6.7 V and 2 = 1.4 V. What is the magnitude of the current in the circuit?
Image of the circuit is at the following link:
http://tinypic.com/r/2ltpwrn/5
Consider the circuit in the figure below with resistors R1 = 2,400 Ω, R2 = 4,400 Ω, and R3 = 1,900 Ω, and batteries with emfs 1 = 6.7 V and 2 = 1.4 V. What is the magnitude of the current in the circuit?
Image of the circuit is at the following link:
http://tinypic.com/r/2ltpwrn/5
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You are close. If ε1 or ε2 were reversed, then you would be right. They would be working together.
Then:
I = (ε1+ε2) / (R1+R2+R3)
But the two +'s are on the same side, which means they are opposing each other. You have to make a small change to your formula. (I = 0.609A).
Then:
I = (ε1+ε2) / (R1+R2+R3)
But the two +'s are on the same side, which means they are opposing each other. You have to make a small change to your formula. (I = 0.609A).
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Look closely at the batteries in your pic. They OPPOSE each other. So instead of summing e1 and e2, what would you do instead? I agree summing the resistances.