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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 \(.....\,\Omega\)

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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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