Question

If the current I through a resistor is increased by 50% then find the increase in power dissipated

Answer 1

When the current in a Resistor is increased by 50% then the Increase in power dissipation is 125 %

Explanation:

Since   P = $I^{2} R$

Current after increased by 50 %

                = I+ $\frac{50}{100} I$ = $\frac{3}{2} I$

     $P^{'} =(\frac{3}{2}I) ^{2} R = \frac{9}{4} I^{2} R$

percentage increase in power dissiaption

$\frac{P^{'}-P }{P} X100$  $\frac{9/4 I^{2}R - I^{2}R }{I^{2} R}$ × 100

N = $\frac{5}{4} x 100 = 125 %$ %