P28_015

P28_015 - part (a) shows. (a) The current in the circuit is...

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15. To be as general as possible, we refer to the individual emf’s as E 1 and E 2 and wait until the latter steps to equate them ( E 1 = E 2 = E ). The batteries are placed in series in such a way that their voltages add; that is, they do not “oppose” each other. The total resistance in the circuit is therefore R total = R + r 1 + r 2 (where the problem tells us r 1 >r 2 ), and the “net emf” in the circuit is E 1 + E 2 . Since battery 1 has the higher internal resistance, it is the one capable of having a zero terminal voltage, as the computation in
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Unformatted text preview: part (a) shows. (a) The current in the circuit is i = E 1 + E 2 r 1 + r 2 + R , and the requirement of zero terminal voltage leads to E 1 = ir 1 = R = E 2 r 1 E 1 r 2 E 1 which reduces to R = r 1 r 2 when we set E 1 = E 2 . (b) As mentioned above, this occurs in battery 1....
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