Easy INTEGER +4 / -1 PYQ · JEE Mains 2021

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)

Answer (integer) 2500

Solution

<p>Given, energy dissipated by a resistor, H = 10 mJ = 10 $\times$ 10<sup>$-$3</sup> J</p> <p>Time, t = 1 s</p> <p>Electric current, I = 2 mA = 2 $\times$ 10<sup>$-$3</sup> A</p> <p>Resistance, R = ?</p> <p>According to Joule's law of heating,</p> <p>H = I<sup>2</sup>Rt</p> <p>$\Rightarrow R = {H \over {{I^2}T}}$ ....... (i)</p> <p>Substituting the given values in Eq. (i), we get</p> <p>$R = {{10 \times {{10}^{ - 3}}} \over {{{(2 \times {{10}^{ - 3}})}^2} \times 1}}$</p> <p>$$ \Rightarrow R = {{{{10}^{ - 2}}} \over {4 \times {{10}^{ - 6}}}} \Rightarrow R = 0.25 \times {10^4}$$</p> <p>$\Rightarrow R = 2500\,\Omega$</p>

About this question

Subject: Physics · Chapter: Current Electricity · Topic: Ohm's Law and Resistance

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