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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

Ask a homework question - tutors are online