Q81 / 90 · JEE Main 2021 (16 Mar Shift 2) · Physics

The energy dissipated by a resistor is $10 mJ$ in $1 s$ , when an electric current of $2 mA$ flows through it. The resistance is ________ $Ω$ . (Round off to the Nearest Integer)

Correct answer: 2500

Solution

$Q = i^{2} R T$

$R = \frac{Q}{i^{2} t} = \frac{10 \times 10^{- 3}}{4 \times 10^{- 6} \times 1} = 2500 Ω$

Hence, the correct answer is 2500.

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