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Question
A cell supplies a current of 0.6 A through a 2Ω coil and a current of 0.3 A through on 8Ω coil. Calculate the e.m.f and internal resistance of the cell.
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Solution
Given: I1 = 0.6 A, R1 = 2 Ω and I2 = 0.3 A, R2 = 8 Ω
Let the e.m.f. of the cell is E and its internal resistance is r.
From I = `"E"/("R + r")`
0.6 = `"E"/("2 + r")` and `0.3 = "E"/("8 + r")`
∴ E = 0.6 (2 + r) = 1.2 + 0.6 r and
E = 0.3 (8 + r) = 2.4 + 0.3r
Thus, 1.2 + 0.6r = 2.4 + 0.3r
or 0.6r - 0.3r = 2.4 - 1.2
Or 0.3 r = 1.2 or
r = `1.2/0.3 = 4 Omega`.
and E = 1.2 + 0.6 × 4 = 1.2 + 2.4 = 3.6 V
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