21. To be as general as possible, we refer to the individual emf’s as ε1and 2and wait until the latter steps to equate them (1= 2= ). 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 Rtotal= R + r1+ r2(where the problem tells us r1> r2), and the“net emf”in the circuit is 1+ 2. Since battery 1 has the higher internal resistance, it is the one capable of having a zero terminal voltage, as the computation in part (a)
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.