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Current Electricity question

2021 · 16 Mar · Shift 2 · Q63
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Current Electricity question

2021 · 16 Mar · Shift 2 · Q63

JEE MainPhysicsCurrent ElectricityNumerical+4 / −1
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 ‾\underline{\hspace{2cm}}​Ω\OmegaΩ. (Round off to the Nearest Integer)
Numerical answer
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Correct answer: 2500

  1. Given data

    • Energy dissipated: E=10 mJ=10×10−3 J=0.01 JE = 10\text{ mJ} = 10 \times 10^{-3}\text{ J} = 0.01\text{ J}E=10 mJ=10×10−3 J=0.01 J
    • Time: t=1 st = 1\text{ s}t=1 s
    • Current: I=2 mA=2×10−3 AI = 2\text{ mA} = 2 \times 10^{-3}\text{ A}I=2 mA=2×10−3 A

  2. Use power-energy relation P=Et=0.011=0.01 WP = \frac{E}{t} = \frac{0.01}{1} = 0.01\text{ W}P=tE​=10.01​=0.01 W

  3. Use Joule heating formula P=I2RP = I^2 RP=I2R Therefore, R=PI2R = \frac{P}{I^2}R=I2P​

  4. Substitute values R=0.01(2×10−3)2R = \frac{0.01}{(2 \times 10^{-3})^2}R=(2×10−3)20.01​ R=0.014×10−6R = \frac{0.01}{4 \times 10^{-6}}R=4×10−60.01​ R=2500 ΩR = 2500\,\OmegaR=2500Ω

  5. Nearest integer R=2500R = 2500R=2500

So, the resistance is 2500 Ω2500\,\Omega2500Ω.

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