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A cell of internal resistance r drives a current through an external resistance R. The power delivered by the cell to the external resistance is maximum when (a) R = r (b) R >> r (c) R << r (d) R = 2r |
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Answer» Correct Answer is: (a) R = r Let ℰ = emf of the cell, i = current in the circuit i = ℰ / R + r. Power delivered to R = P = i2R = ℰ2R/(R + r)2 = f(R). For P to be maximum, dP/dR = 0 = ℰ2 [1 / (R + r)2 - 2R / (R + r)3] or 2R = R + r or R = r |
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