Question

Two cells of same emf E but internal resistance r₁ and r₂ are connected in series to an external resistor R (Figure). What should be the value of R so that the potential difference across the terminals of the first cell becomes zero.

Answer 1

In this problem first we apply Ohm' law to find current in the circuit.

Effective  emf of two cells = E + E = 2E

Effective resistance = R + r₁ + r²

So the electric current is given by

`I = (E + E)/(R + r_1 + r_2)`

`V_1 = E - Ir_1 = E - (2E.r_1)/(R + r_1 + r_2)`

The net potential difference across 1^st cell V₁ = 0 (Given)

∴ `E - (2Er_1)/(R + r_1 + r_2)` = 0

Or `1 - (2r_1)/(R + r_1 + r_2)` = 0

`(2r_1)/(r_1 + r_2 + R) = 1/1`

Or 2r₁ = r₁ + r₂ + R

r1 – r2 = R

It is the required condition for the potential difference across 1^st cell to be zero.