Q87 / 90 · JEE Main 2019 (10 Jan Shift 2) · Physics

A current of $2 mA$ was passed through an unknown resistor which dissipated a power of $4.4 W .$ Dissipated power when an ideal power supply of $11 V$ is connected across it is:

A
$11 \times 10^{5} W$
B
$11 \times 10^{- 3} W$
Correct answer
C
$11 \times 10^{- 5} W$
D
$11 \times 10^{- 4} W$

Solution

$P = \frac{V^{2}}{R} = i^{2} R R e s i s t a n c e R = \frac{V^{2}}{P} = \frac{P}{i^{2}}$
Here, Power $P = 4.4 W a n d i = 2 mA$

so $R = \frac{4.4}{4 \times 10^{- 6}}$ $Ω$

When $11 V$ power supply is connected across R

$P^{'} = \frac{11^{2}}{4.4} \times 4 \times 10^{- 6}$

$= 11 \times 10^{- 5} W$ .

Hence, the correct answer is option: 'C'.

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A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11 V is connected across it is:

Solution

Relevant Questions

A current of $2 mA$ was passed through an unknown resistor which dissipated a power of $4.4 W .$ Dissipated power when an ideal power supply of $11 V$ is connected across it is: