Question:

$ A\,220\,V,100\,W $ bulb is joined with a $110\, V$ supply. The power consumed by the bulb is

Updated On: Jul 12, 2022
  • 50 W
  • 25 W
  • 80 W
  • 100 W
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The Correct Option is B

Solution and Explanation

The electric power $P$ is given by $P=\frac{W}{t}=\frac{V^{2}}{R} $ watt where, $V$ is potential difference and $R$ the resistance. Given, $ P_{1} =100 \,W , V_{1}=220\, V $ $ \therefore R =\frac{V_{1}^{2}}{P_{1}}=\frac{(220)^{2}}{100} $ $=484 \,\Omega $ Hence, power dissipated when potential difference is $110 \,V$ is $P=\frac{(110)^{2}}{484}=25 \,W$
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Top Questions on Electromagnetic induction

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Concepts Used:

Electromagnetic Induction

Electromagnetic Induction is a current produced by the voltage production due to a changing magnetic field. This happens in one of the two conditions:-

  1. When we place the conductor in a changing magnetic field.
  2. When the conductor constantly moves in a stationary field.

Formula:

The electromagnetic induction is mathematically represented as:-

e=N × d∅.dt

Where

  • e = induced voltage
  • N = number of turns in the coil
  • Φ = Magnetic flux (This is the amount of magnetic field present on the surface)
  • t = time

Applications of Electromagnetic Induction

  1. Electromagnetic induction in AC generator
  2. Electrical Transformers
  3. Magnetic Flow Meter