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Current Electricity question

2019 · 10 Jan · Shift 2 · Q67
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Current Electricity question

2019 · 10 Jan · Shift 2 · Q67

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
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 -
  1. A
    11 ×\times× 10–5 W
  2. B
    11 ×\times× 10–3 W
  3. C
    11 ×\times× 105 W
  4. D
    11 ×\times× 10–4 W
View written solutionFree

Correct answer: A

  1. Given data
  • Current through the unknown resistor:
    I=2 mA=2×10−3 AI = 2\text{ mA} = 2 \times 10^{-3}\text{ A}I=2 mA=2×10−3 A
  • Power dissipated:
    P=4.4 WP = 4.4\text{ W}P=4.4 W

We first find the resistance of the resistor.

  1. Use power relation for a resistor

For a resistor, P=I2RP = I^2 RP=I2R

So, R=PI2R = \frac{P}{I^2}R=I2P​

Substitute the values: R=4.4(2×10−3)2R = \frac{4.4}{(2 \times 10^{-3})^2}R=(2×10−3)24.4​

Now, (2×10−3)2=4×10−6(2 \times 10^{-3})^2 = 4 \times 10^{-6}(2×10−3)2=4×10−6

Hence, R=4.44×10−6=1.1×106 ΩR = \frac{4.4}{4 \times 10^{-6}} = 1.1 \times 10^6\,\OmegaR=4×10−64.4​=1.1×106Ω

  1. Power when 11 V is applied

Now an ideal source of voltage V=11V = 11V=11 V is connected across the same resistor.

Power dissipated is: P′=V2RP' = \frac{V^2}{R}P′=RV2​

Substitute values: P′=1121.1×106P' = \frac{11^2}{1.1 \times 10^6}P′=1.1×106112​

P′=1211.1×106=110×10−6P' = \frac{121}{1.1 \times 10^6} = 110 \times 10^{-6}P′=1.1×106121​=110×10−6

P′=1.1×10−4 WP' = 1.1 \times 10^{-4}\text{ W}P′=1.1×10−4 W

This can be written as: P′=11×10−5 WP' = 11 \times 10^{-5}\text{ W}P′=11×10−5 W

  1. Match with options
  • A: 11×10−5 W11 \times 10^{-5}\text{ W}11×10−5 W ✅
  • B: 11×10−3 W11 \times 10^{-3}\text{ W}11×10−3 W
  • C: 11×105 W11 \times 10^{5}\text{ W}11×105 W
  • D: 11×10−4 W11 \times 10^{-4}\text{ W}11×10−4 W

So the correct option is A.

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