MHT CET · Physics · Alternating Current
In an \(L-R\) circuit, the inductive reactance is equal to \(\sqrt{3}\) times the resistance ' \(R\) ' of the circuit. An e.m.f. \(E=E_0 \sin (\omega t)\) is applied to the circuit. The power consumed in the circuit is
- A \(\frac{E_0^2}{4 R}\)
- B \(\frac{E_0^2}{6 R}\)
- C \(\frac{E_0^2}{8 R}\)
- D \(\frac{E_0^2}{12 R}\)
Answer & Solution
Correct Answer
(C) \(\frac{E_0^2}{8 R}\)
Step-by-step Solution
Detailed explanation
\(Z = \sqrt{R^2 + X_L^2}\) \(Z = \sqrt{R^2 + (\sqrt{3}R)^2} = \sqrt{R^2 + 3R^2} = \sqrt{4R^2} = 2R\)
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