Q148 / 160 · AP EAMCET 2022 (08 Jul Shift 2) · Physics

A heating element is designed to dissipate $2400 \mathrm{~W}$ when connected to $240 \mathrm{~V}$. The power it dissipates when it is connected to $120 \mathrm{~V}$ is (Assume that resistance of the filament is constant)

  • Correct answer
    A
    $600 \mathrm{~W}$
  • B
    $1200 \mathrm{~W}$
  • C
    $1800 \mathrm{~W}$
  • D
    $400 \mathrm{~W}$

Solution

Given, Rated power, $P_R=2400 \mathrm{~W}$ Rated voltage, $V_R=240 \mathrm{~V}$ So, resistance of element, $R=\frac{V_R^2}{P_R} \quad\left(\because P=\frac{V^2}{R}\right)$ $=\frac{240 \times 240}{2400}=24 \Omega$ when element is connected across $120 \mathrm{~V}$ supply dissipated power is, $\begin{aligned} P & =\frac{V^2}{R}=\frac{120 \times 120}{24} \\ & =600 \mathrm{~W}\end{aligned}$

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

Relevant Questions