Q114 / 150 · MHT CET 2021 (22 Sep Shift 2) · Physics

For the circuit shown below, instantaneous current through inductor ' $\mathrm{L}$ ' and capacitor ' $\mathrm{C}$ ' is respectively. Figure

  • Correct answer
    A
    $\frac{-\mathrm{e}_0}{\omega \mathrm{L}} \cos \omega \mathrm{t} ; \mathrm{e}_0 \omega \mathrm{c} \cos \omega \mathrm{t}$
  • B
    $\frac{-\mathrm{e}_0}{\omega \mathrm{L}} \sin \omega \mathrm{t} ; \frac{\mathrm{e}_0}{\omega \mathrm{C}} \cos \omega \mathrm{t}$
  • C
    $\frac{\mathrm{e}_0 \mathrm{C}}{\mathrm{L}} \cos \omega \mathrm{t} ; \frac{\mathrm{e}_0 \mathrm{~L}}{\mathrm{C}} \sin \omega \mathrm{t}$
  • D
    $\frac{-\mathrm{e}_0 \mathrm{C}}{\mathrm{L}} \sin \omega \mathrm{t} ; \frac{\mathrm{e}_0 \mathrm{~L}}{\mathrm{C}} \cos \omega \mathrm{t}$

Solution

The current through the inductor lags the applied emf by $\frac{\pi}{2}$ and the current through the capacitor leads the current by $\frac{\pi}{2}$ $$ \begin{aligned} & \therefore \mathrm{i}_{\mathrm{L}}=\frac{\mathrm{e}_0}{\omega \mathrm{L}} \sin \left(\omega \mathrm{t}-\frac{\pi}{2}\right)=-\frac{\mathrm{e}_0}{\omega \mathrm{L}} \cos \omega \mathrm{t} \\ & \text { and } \mathrm{i}_{\mathrm{c}}=\mathrm{e}_0 \omega \mathrm{C} \sin \left(\omega \mathrm{t}+\frac{\pi}{2}\right)=\mathrm{e}_0 \omega \mathrm{C} \cos \omega \mathrm{t} \end{aligned} $$

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

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