An AC voltage is applied to a resistance $R$ and an inductor $L$ in series. If $R$ and the inductive…

An AC voltage is applied to a resistance $R$ and an inductor $L$ in series. If $R$ and the inductive reactance are both equal to $3 \Omega$, the phase difference between the applied voltage and the current in the circuit is
  1. $\pi / 4$
  2. $\pi / 2$
  3. zero
  4. $\pi / 6$

Solution

$\begin{aligned} \tan \phi=\frac{X_L}{R} & =\frac{L \omega}{R} \\ \tan \phi & =\frac{3 \Omega}{3 \Omega} \\ \tan \phi & =1 \\ \phi & =\tan ^{-1}(1) \\ \phi & =45^{\circ} \\ \phi & =\frac{\pi}{4} \mathrm{rad} \end{aligned}$

Asked in: MHT CET Full Test 13

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