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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 Resistnce of the Cell. - Physics

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प्रश्न

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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उत्तर

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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Internal Resistance of a Cell
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अध्याय 7: Electricity - Short Numericals

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आईसीएसई Physics [English] Class 10
अध्याय 7 Electricity
Short Numericals | Q 1

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