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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 |
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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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