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Question
Study the two circuits shown in the figure below. The cells in the two circuits are identical to each other. The resistance of the load resistor R is the same in both circuits.
If the same current flows through the resistor R in both circuits, calculate the internal resistance of each cell in terms of the resistance of resistor R. Show your calculations.
Solution
Current I in load resistance R is given by the formula,
I = `"E"/("R"+"r")`;
where E is the emf of the cell,
r is the internal resistance of the cell.
In circuit 1, the cells are in series.
∴ Eeq = 2E and req = 2r
Current through R in circuit 1, I = `(2"E")/(("R"+2"r"))`
In circuit 2, the cells are in parallel.
∴ Eeq = E and req = `"r"/2`
Current through R in circuit 2, I = `"E"/(("R"+"r"/2))`
Since the current is the same in both circuits,
`(2"E")/ (("R"+2"r")) = "E"/(("R"+"r"/2))`
Solving, we get r = R
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