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I have a question from a workbook:

A storage battery of emf (Ve) 34V and internal resistance 0.1Ω is to be charged at a rate of 20A from a 110-V source. What series resistance is needed in the circuit?

I used formula Ve = I(R + r): 34V = (20A)(R + 0.1Ω) to find that R = 1.59Ω. I think this means that if there were a series circuit with just Ve and a resistor, then the resistor would have to be 1.59Ω in order for current to be 20A. So if we add 110-V source into the circuit, then the resistance changes while current remains as 20A. I assume the formula V = IR would come into play...but I'm not sure how to proceed. What is the next step in solving this problem?

By the way, the solution was given as 3.7Ω. I'd appreciate any insight :)

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You are right that the total resistance is $R + r$. You also know that the total current is $20A$, and the total voltage is $110V - 34V$. Apply Ohm's law to these three values and solve for $R$. (I'll leave it to you to explain why you subtract the two voltages rather than adding them.)

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  • $\begingroup$ Oh, I see how to get 3.7Ω... $\endgroup$ – J.K Oct 17 '16 at 6:53
  • $\begingroup$ So the voltage drop of the emf is 2V and the voltage drop of resistor is 74 and they add to get 76V. Therefore, that is the total voltage? Doesn't the 110V source also have an internal resistance though? $\endgroup$ – J.K Oct 17 '16 at 6:55
  • $\begingroup$ Yes, any real source will have some internal resistance, but if it's not mentioned in the problem statement, I guess you are allowed to assume it is negligible in this case. $\endgroup$ – Paul G Oct 17 '16 at 13:55

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