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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.

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प्रश्न

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.

संक्षेप में उत्तर
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उत्तर

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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