An inductance of $\frac{300}{\pi} \mathrm{mH}$, a capacitance of $\frac{1}{\pi} \mathrm{mF}$ and a…

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
  1. $\tan ^{-1}(0)$
  2. $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
  3. $\tan ^{-1}(1)$
  4. $\tan ^{-1}(\sqrt{3})$

Solution

$\begin{aligned} & \tan \phi=\frac{X_L-X_C}{R}=\left(\frac{\omega L-\frac{1}{\omega C}}{R}\right) \\ & =\frac{(2 \times \pi \times 50) \times\left(\frac{300}{\pi} \times 10^{-3}\right)-\frac{1}{(2 \times \pi \times 50) \times\left(\frac{1}{\pi} \times 10^{-3}\right)}}{20} \end{aligned}$ i.e. $\tan \phi=1$ $\phi=\tan ^{-1}(1)$

Asked in: MHT CET 2024 (11 May Shift 1)

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