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Two batteries of emf E1 and E2, with internal resistances r1 and r2, are connected with a resistance R. Under what condition does the current through R increase when the battery of emf E2 is short-circuited? (E1 drives the main current; r1 and r2 are the internal resistances.)

1. E2*r1 > E1*(R + r2)
2. E1*r2 > E2*(R + r1)
3. E2*r2 > E1*(R + r2)
4. E1*r1 > E2*(R + r1)

Correct answer: E2*r1 > E1*(R + r2)

Solution

Writing the loop current with both cells acting versus with E2 shorted (its branch becoming just r2) and demanding the latter exceeds the former leads to the condition E2*r1 > E1*(R + r2).

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