Q164 / 170 · Manipal (MET) 2020 · Physics

On a heater coil it is written that, 250 V , 500 W . What is the resistance of this coil?

  • A
    $62.5 \Omega$
  • B
    $100 \Omega$
  • C
    $200 \Omega$
  • Correct answer
    D
    $125 \Omega$

Solution

Given $V=250 \mathrm{~V}$ and $P=500 \mathrm{~W}$ As we know that, $P=\frac{V^2}{R}$ $\Rightarrow \quad R=V^2 / P=\frac{250 \times 250}{500}$ $=\frac{625}{5}=125 \Omega$

Hence, the correct answer is option: 'D'.

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