COMEDK · Physics · 23. Alternating Current
An inductor \(1 \mathrm{H}\) is connected across \(220 \mathrm{~V}\), \(50 \mathrm{~Hz}\) supply. Peak value of current is approximately
- A \(0.5 \mathrm{~A}\)
- B \(0.7 \mathrm{~A}\)
- C \(1 \mathrm{~A}\)
- D \(1.4 \mathrm{~A}\)
Answer & Solution
Correct Answer
(C) \(1 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
Given, \(L=1 \mathrm{H}, V=220 \mathrm{~V}, f=50 \mathrm{~Hz}\) Peak value of current \(I_{0}=\frac{V_{0}}{X_{L}}=\frac{V_{0}}{\omega L}\) \(\Rightarrow I_{0}=\frac{\sqrt{2} V}{2 \pi f L}=\frac{\sqrt{2} \times 220}{2 \times \pi \times 50 \times 1} \simeq 1 \mathrm{~A}\)
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