Question:
An electric bulb rated as $200 \mathrm{~W}$ at $100 \mathrm{~V}$ is used in a circuit having $200 \mathrm{~V}$ supply. The resistance ' $R$ ' that must be put in series with the bulb so that the bulb delivers the same power is________$\Omega$
Solution:
Power, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{B}}}$
$\mathrm{R}_{\mathrm{B}}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{100 \times 100}{200}$
$\mathrm{R}_{\mathrm{B}}=50 \Omega$
To produce same power, same voltage (i.e. $100 \mathrm{~V}$ ) should be across the bulb
Hence, $R=R_{B}$
$R=50 \Omega$
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