Question

Figures 4 and 5 represent the combination of two identical cells having negligible internal resistance.

1. In which of the two combinations, emf of the battery is greater?

2. When a resistor ‘R’ is connected between the terminals P and Q, I₁ and I₂ are the currents flowing through ‘R’ in Figure 4 and Figure 5 respectively. Obtain the ratio `I_1/I_2`.

Answer 1

1. For a series combination, Net emf (E_S) = E + E = 2E, and for a parallel combination, Net emf is equal to the emf of a single cell. (E_P) = E
2. `I_1/I_2 = (2E//R)/(E//R) = 2/1`    ...`[because I = "emf"/"Resistance"]`