MHT CET · Physics · Alternating Current
An inductance coil has a resistance of \(100 \Omega\). When are a.c. signal of frequency \(100 \mathrm{Hzis}\) applied to the coil, the voltage leads the current by \(45^{\circ}\). The inductance of the coil in henry is \(\left[\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right]\)
- A \(\frac{1}{\pi}\)
- B \(\frac{5}{2 \pi}\)
- C \(\frac{2}{\pi}\)
- D \(\frac{1}{2 \pi}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{2 \pi}\)
Step-by-step Solution
Detailed explanation
The phase angle in an LR circuit is given by, since \(\phi=45^{\circ}\),
\(\tan \phi=\frac{X_L}{X_R}=1\)
Resistance of inductive coil \(X_L=X_R=100 \Omega\)
Thus, inductance of coil \(L=\frac{X_L}{2 \pi f}\)
\(\therefore L=\frac{100}{2 \pi \times 100 \mathrm{~s}^{-1}}=\frac{1}{2 \pi} \mathrm{H}\)
\(\tan \phi=\frac{X_L}{X_R}=1\)
Resistance of inductive coil \(X_L=X_R=100 \Omega\)
Thus, inductance of coil \(L=\frac{X_L}{2 \pi f}\)
\(\therefore L=\frac{100}{2 \pi \times 100 \mathrm{~s}^{-1}}=\frac{1}{2 \pi} \mathrm{H}\)
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