JEE Mains · Physics · STD 12 - 3. current electricity
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 \(.....\,\Omega\)
- A \(15\)
- B \(20\)
- C \(5\)
- D \(50\)
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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