P28_015.PDF

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

). 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 =

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

part (a) shows.

E 1 +

(a) The current in the circuit is

E 2

r 1 + r 2 + R ,

and the requirement of zero terminal voltage leads to

i = E 1 +

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.

equate them (