JEE Mains · Physics · STD 12 -7. Alternating current
An electric bulb rated as \(100 \mathrm{~W}-220 \mathrm{~V}\) is connected to an ac source of rms voltage 220 V. The peak value of current through the bulb is :
- A \(0.64\mathrm{~A}\)
- B \(0.45\mathrm{~A}\)
- C \(2.2\mathrm{~A}\)
- D \(0.32\mathrm{~A}\)
Answer & Solution
Correct Answer
(A) \(0.64\mathrm{~A}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{P}=\mathrm{v}_{\mathrm{rms}} \mathrm{i}_{\mathrm{rms}} \\ & \mathrm{i}_{\mathrm{rms}}=\frac{100}{220} \\ & \mathrm{i}_0=\sqrt{2} \mathrm{i}_{\mathrm{rms}}=0.64 \mathrm{~A}\end{aligned}\)
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