JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2021
- Medium
- Numerical
An electric bulb rated as \(200\, {W}\) at \(100 \,{V}\) is used in a circuit having \(200\, {V}\) supply. The resistance \('R'\) that must be put in series with the bulb so that the bulb delivers the same power is \(.....\,\Omega\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(50\)
Step-by-step Solution
Detailed explanation
Power, \(P =\frac{ V ^{2}}{ R _{ B }}\)
\(R_{B}=\frac{V^{2}}{P}=\frac{100 \times 100}{200}\)
\(R_{B}=50 \Omega\)
To produce same power, same voltage should be across the bulb
Hence, \(R=R_{B}=50 \Omega\)
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