AP EAMCET · PHYSICS · Alternating Current
When a coil is connected to AC supply of frequency \(50 \mathrm{~Hz}\), a current of \(4 \mathrm{~A}\) flows in it and it consumes \(240 \mathrm{~W}\) power. If the potential difference across the coil is \(100 \mathrm{~V}\), then the inductance value of the coil is
- A \(\mathrm{L}=(5 \pi) \mathrm{H}\)
- B \(L=\frac{\pi}{5} \mathrm{H}\)
- C \(L=\frac{1}{5 \pi} \mathrm{H}\)
- D \(L=\frac{1}{25 \pi} \mathrm{H}\)
Answer & Solution
Correct Answer
(C) \(L=\frac{1}{5 \pi} \mathrm{H}\)
Step-by-step Solution
Detailed explanation
\(I^2 R=240 \mathrm{~W}, I=4 \mathrm{~A}\) \[ \therefore \quad R=\frac{240}{16}=15 \Omega \] As, \(V=I Z=I \sqrt{X_L^2+R^2} \Rightarrow \frac{V}{I}=\sqrt{X_L^2+R^2}\)…
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