JEE PYQ: Current Electricity Question 4
Question 4 - 2021 (16 Mar 2021 Shift 2)
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 ____ $\Omega$. (Round off to the Nearest Integer)
Show Answer
Answer: 2500
Solution
$Q = i^2 R t$. $R = \frac{Q}{i^2 t} = \frac{10 \times 10^{-3}}{(2 \times 10^{-3})^2 \times 1} = \frac{10 \times 10^{-3}}{4 \times 10^{-6}} = 2500,\Omega$.
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