An $R-L-C$ circuit consists of a $150 \Omega$ resistor, $20 \mu \mathrm{F}$ capacitor and a $500…

An $R-L-C$ circuit consists of a $150 \Omega$ resistor, $20 \mu \mathrm{F}$ capacitor and a $500 \mathrm{mH}$ inductor connected in series with a $100 \mathrm{~V}$ AC supply. The angular frequency of the supply voltage is $400 \mathrm{rad} \mathrm{s}^{-1}$. The phase angle between current and the applied voltage is
  1. $\tan ^{-1}(0.8)$
  2. $\tan ^{-1}(0.25)$
  3. $\tan ^{-1}(0.6)$
  4. $\tan ^{-1}(0.5)$

Solution

In $R-L-C$ series circuit $\begin{aligned} R=150 \Omega, C & =20 \mu \mathrm{F} \\ & =2 \times 10^{-5} \mathrm{~F}\end{aligned}$ $L=500 \mathrm{mH}=0.5 \mathrm{H}$ Supply voltage, $V=100 \mathrm{~V}$ Angular frequency, $\omega=400 \mathrm{rads}^{-1}$ Now, $X_L=\omega L$ $=400 \times 0.5=200 \Omega$ $X_C=\frac{1}{\omega C}=\frac{1}{400 \times 2 \times 10^{-5}} \Rightarrow X_C=125 \Omega$ $X_L>X_C$ $\therefore \tan \phi=\frac{X_L-X_C}{R}=\frac{200-125}{150}=\frac{75}{150}=0.5$ $\therefore \quad \phi=\tan ^{-1}(0.5)$

Asked in: AP EAMCET 2022 (05 Jul Shift 1)

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