An electric motor runs on a d.c. source of emf `epsilon` and internal resistance r. show that the power output of the source is maximum when the current drawn by the motor is `epsilon//2r`.
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Let `I` be the current drawn by motor. Then power output of the source, `P = epsilon I - I^(2) r` P will be maximum, when `(dP)/(dI) = 0 or (d)/(dt) (epsilon I - I^(2)r) =0` or `epsilon - 2 Ir = 0 or I = (epsilon)/(2r)`
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