JEE Advanced · PhysicsHard

$A battery of emf E and internal resistance r is connected to a variable resistor R . The power delivered to the resistor is maximum when R = r . If the resistor is replaced by a network of resistors such that the equivalent resistance is R, what is the maximum power dissipated in the network?$

Show correct answer & step-by-step solution

Correct answer: A — $P_{max} = \frac{E ^{2}}{4 r}$

Solution

$The current in the circuit is I = \frac{E}{R + r} .$
$The power dissipated in the load is P = I^{2} R = \frac{E ^{2} R}{( R + r ) ^{2}} .$
$To find the maximum, set \frac{d P}{d R} = 0, which yields R = r .$
$Substituting R = r into the power formula gives P_{max} = \frac{E ^{2} r}{( r + r ) ^{2}} = \frac{E ^{2}}{4 r} .$
$Hence the answer is (A).$