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 :

Engineering

Physics

Ohm Law and Current Basics

Question

11 × 10^–4 W

11 × 10^–5 W

11 × 10⁵ W

11 × 10^–3 W

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$R = \frac{R}{I^{2}} = \frac{4 .4}{4 \times 10^{- 6}} Ω$

$P' = \frac{V^{2}}{R} = \frac{11 \times 11 \times 4 \times 10^{- 6}}{4 .4} W$

= 11 × 10^–5 W