Current Electricity

JEE Main level, single correct, medium

257 of 380

Question 257 of 380

A current of 2mA2 mA was passed through an unknown resistor which dissipated a power of 4.4W.4.4W. Dissipated power when an ideal power supply of 11V11 V is connected across it is:

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Answer: (C) 11×10−5W11\times {10}^{-5}W

P=V2R=i2RResistanceR=V2P=Pi2P=\dfrac{{V}^{2}}{R}={i}^{2}R ResistanceR=\dfrac{{V}^{2}}{P}=\dfrac{P}{{i}^{2}}
Here, Power P=4.4Wandi=2mAP=4.4Wandi=2mA

so R=4.44×10−6R=\dfrac{4.4}{4 \times {10}^{-6}} Ω\Omega

When 11V11V power supply is connected across R

P′=1124.4×4×10−6{P}^{'}=\dfrac{{11}^{2}}{4.4}\times 4\times {10}^{-6}

=11×10−5W=11\times {10}^{-5}W.

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