Q127 / 155 · WBJEE 2022 · Physics

When an $A C$ source of emf $E$ with frequency $\omega=100 \mathrm{~Hz}$ is connected across a circuit, the phase difference between $E$ and current I in the circuit is observed to be $\frac{\pi}{4}$ as shown in the figure. If the circuit consist of only $\mathrm{RC}$ or $\mathrm{RL}$ in series, then Figure

  • A
    $\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=5 \mu \mathrm{F}$
  • B
    $\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{L}=10 \mathrm{H}$
  • C
    $\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{L}=1 \mathrm{H}$
  • Correct answer
    D
    $\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=10 \mu \mathrm{F}$

Solution

.. Since supply voltage lags the current $\operatorname{By}(\phi=\pi / 4)$ $$ \begin{aligned} &\tan \phi=\frac{\mathrm{X}_{\mathrm{C}}}{\mathrm{R}}=1, \mathrm{X}_{\mathrm{C}}=\mathrm{R} \ldots \ldots(1) \\ &\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{100 \times 10 \times 10^{-6}}=1 \mathrm{k} \Omega=\mathrm{R} \end{aligned} $$

Hence, the correct answer is option: 'D'.

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