MHT CET · Physics · Alternating Current
An inductance of \(\frac{300}{\pi} \mathrm{mH}\), a capacitance of \(\frac{1}{\pi} \mathrm{mF}\) and a resistance of \(20 \Omega\) are connected in series with an a.c. source of \(240 \mathrm{~V}, 50 \mathrm{~Hz}\). The phase angle of the circuit is
- A \(\tan ^{-1}(0)\)
- B \(\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
- C \(\tan ^{-1}(1)\)
- D \(\tan ^{-1}(\sqrt{3})\)
Answer & Solution
Correct Answer
(C) \(\tan ^{-1}(1)\)
Step-by-step Solution
Detailed explanation
\(\tan \phi=\frac{X_L-X_C}{R}=\left(\frac{\omega L-\frac{1}{\omega C}}{R}\right)\)
\(=\frac{(2 \times \pi \times 50) \times\left(\frac{300}{\pi} \times 10^{-3}\right)-
\frac{1}{(2 \times \pi \times 50) \times\left(\frac{1}{\pi} \times 10^{-3}\right)}}{20}\)
i.e. \(\tan \phi=1\)
\(\phi=\tan ^{-1}(1)\)
\(=\frac{(2 \times \pi \times 50) \times\left(\frac{300}{\pi} \times 10^{-3}\right)-
\frac{1}{(2 \times \pi \times 50) \times\left(\frac{1}{\pi} \times 10^{-3}\right)}}{20}\)
i.e. \(\tan \phi=1\)
\(\phi=\tan ^{-1}(1)\)
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