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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.
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Solution
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 1st 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 1st cell to be zero.
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