e.m.f. problem?

Question

Two batteries os e.m.f.'s E1 and E2 and internal resistances r1 and r2 are connected in parallel.Their equivalent e.m.f. will be
1. E1r1+E2r2\E1+E2
2. E1r2+E2r1\r1+r2
3. r1+r2\E1r1+E2r2
4. r1+r2\E1r2+E2r1

Answer 1

Think of the setup as two ideal batteries connected together directly at one terminal and connected together via a voltage divider at the other. Then the voltage you are looking for is the voltage at the junction of the two divider resistors:
E1 + (E2 - E1) * (r1 / (r1 + r2))
You can do the algebra. (Hint: The 'most correct' choice needs a pair of parentheses.)