Two battery hooked up in series, energy conversion

[Helmholtz]

Homework Statement

The positive terminals of two batteries with emf's of V_1 and V_2, respectively, are connected together. It is V_2>V_1. The circuit is completed by connecting the negative terminals. If each battery has an internal resistance r, what is the rate with which electrical energy is converted to chemical energy in the smaller battery?

Answer: (V_2-V_1)*V_1/r

Homework Equations

P=IV
V=IR

The Attempt at a Solution

From what I was trying is that I=(V_2-V_1)/(2r) and then the voltage difference is V_1, so I was getting P= (V_2-V_1)*V_1/(2r), what am I doing wrong that I'm off back a factor of two.

 

[lewando]

Your answer looks right (official answer looks wrong), but your reasoning is not clear.

...and then the voltage difference is V_1, so I was getting P= (V_2-V_1)*V_1/(2r)

Your expression for I is right. Think of these three things:
1) the energy source is the ideal portion of the real battery with emf of V_2.
2) the two internal resistors are sinking energy, actually removing energy from the circuit.
3) the ideal portion of the real battery emf of V_1 is also an energy sink but is storing energy.

Also the answer is the initial rate-- the steady state rate is zero.

 

[CWatters]

I agree the book answer looks wrong.

I think your reasoning is fine but your explanation could be slightly better.

Perhaps point out that "the rate with which electrical energy is converted to chemical energy" is equivalent to the the power dissipated in the ideal voltage V_1.

 

[rude man]

I also think the OP's answer is right. Energy charging battery 1 is V1*i*t so rate is V1*i. But i = (V2-V1)/2r.