JEE Mains · Physics · STD 12 - 3. current electricity
A \(220 \; V , 50 \; Hz\) AC source is connected to a \(25 \; V\), \(5 \; W\) lamp and an additional resistance \(R\) in series (as shown in figure) to run the lamp at its peak brightness, then the value of \(R\) (in ohm) will be

- A \(975\)
- B \(875\)
- C \(775\)
- D \(675\)
Answer & Solution
Correct Answer
(A) \(975\)
Step-by-step Solution
Detailed explanation
\(P = Vi\) \(5=25 i\) \(i=\frac{1}{5}\) \(V_{R}=i R\) \((220-25)=\frac{1}{5} R\) \(R=195 \times 5=975 \Omega\)
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