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Question
When a resistance of 3Ω is connected across a cell, the current flowing is 0.5 A. On changing the resistance to 7Ω, the current becomes 0.25A. Calculate the e.m.f. and the internal resistance of the cell.
Solution
I = `"E"/("R + r")`
∴ `0.5 = "E"/(3 + "r")` ....(as I = 0.5 A when R = 3Ω)
∴ E = (3 + r) × 0.5 ....(i)
also 0.25 = `"E"/(7 + "r")` ...(as I = 0.25 A when R = 7 W)
∴ E = (7 + r) × 0.25 ....(ii)
Comparing equation (i) and (ii)
E = (3 + r) × 0.5
= (7 + r) × 0.25
∴ 1.5 + 0.5r = 1.75 + 0.25r
∴ 0.25 r = 0.25
Internal resistance = r = 1 Ω
Now E = I (R + r)
∴ E = 0.5(3 + 1) (as I = 0.5 A when R = 3 Ω, r = 1 Ω)
∴ E = 0.5 + 1.5 = 2 V
E.m.f. of cell E = 2 volt = 2 V.
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