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(1) 10 Ω(3) 109 ΩA 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) 109 Ω(2)0.1620.91 Ω(2)0.1 Ω(4)091 Ω |
Answer» <p>From Ohm’s Law</p><p>i = E / (R + r)</p><p>1.1 = 12 / (10 + r)</p><p>11 + 1.1r = 12</p><p>1.1r = 1</p><p>r = 1 / 1.1</p><p>r = 10/11 Ω</p><p>∴ Internal resistance of battery is 10/11 = 0.91 Ω</p> | |