An electric bulb rated 50 ~W-200 ~V is connected across a 100 ~V supply. The power dissipation of the bulb is:

Correct answer: B. 12.5 ~W Rated power \& voltage gives resistance aligned & R= V^2 P =((200)^2)/(50)=(40000)/(50) & R=800 & P= (V_ applied )^2 R =((100)^2)/(800) & P=12.5 watt aligned

JEE Main 20244 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Evening Shift
Subject: Physics
Chapter: Current Electricity

JEE Main 2024 · 4 April · Evening Shift · Physics · Current Electricity

An electric bulb rated 50 ~W-200 ~V is connected across a 100 ~V...

An electric bulb rated

50 \mathrm{~W}-200 \mathrm{~V}

is connected across a

100 \mathrm{~V}

supply. The power dissipation of the bulb is:

Options

A $25 \mathrm{~W}$
B $12.5 \mathrm{~W}$
C $50 \mathrm{~W}$
D $100 \mathrm{~W}$

Correct answer

B. $12.5 \mathrm{~W}$

Step-by-step solution

Rated power \& voltage gives resistance

\begin{aligned} & \mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(200)^2}{50}=\frac{40000}{50} \\ & \mathrm{R}=800 \\ & \mathrm{P}=\frac{\left(\mathrm{V}_{\text{applied }}\right)^2}{\mathrm{R}}=\frac{(100)^2}{800} \\ & \mathrm{P}=12.5 \text{ watt } \end{aligned}

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