1.

A resistor of 10 Ω connected across a cell of emf 12 V, draws the current of 1.1 A.Find the internal resistance of the cell(1) 10 Ω(3) 10-9 Ω(2)0.1 Ω(4)0.91 Ω

Answer»

From Ohm’s Law

i = E / (R + r)

1.1 = 12 / (10 + r)

11 + 1.1r = 12

1.1r = 1

r = 1 / 1.1

r = 10/11 Ω

∴ Internal resistance of battery is 10/11 =0.91 Ω

tnks 0.91 is right answer



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