MHT CET · Physics · Alternating Current
For the circuit shown below, instantaneous current through inductor ' \(\mathrm{L}\) ' and capacitor ' \(\mathrm{C}\) ' is respectively.
- 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 \(\frac{-\mathrm{e}_0 \mathrm{C}}{\mathrm{L}} \sin \omega \mathrm{t} ; \frac{\mathrm{e}_0 \mathrm{~L}}{\mathrm{C}} \cos \omega \mathrm{t}\)
Answer & Solution
Correct Answer
(A) \(\frac{-\mathrm{e}_0}{\omega \mathrm{L}} \cos \omega \mathrm{t} ; \mathrm{e}_0 \omega \mathrm{c} \cos \omega \mathrm{t}\)
Step-by-step Solution
Detailed explanation
The current through the inductor lags the applied emf by \(\frac{\pi}{2}\) and the current through the capacitor leads the current by \(\frac{\pi}{2}\)
\(
\begin{aligned}
& \therefore \mathrm{i}_{\mathrm{L}}=\frac{\mathrm{e}_0}{\omega \mathrm{L}} \sin \left(\omega \mathrm{t}-\frac{\pi}{2}\right)=-\frac{\mathrm{e}_0}{\omega \mathrm{L}} \cos \omega \mathrm{t} \\
& \text { and } \mathrm{i}_{\mathrm{c}}=\mathrm{e}_0 \omega \mathrm{C} \sin \left(\omega \mathrm{t}+\frac{\pi}{2}\right)=\mathrm{e}_0 \omega \mathrm{C} \cos \omega \mathrm{t}
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \mathrm{i}_{\mathrm{L}}=\frac{\mathrm{e}_0}{\omega \mathrm{L}} \sin \left(\omega \mathrm{t}-\frac{\pi}{2}\right)=-\frac{\mathrm{e}_0}{\omega \mathrm{L}} \cos \omega \mathrm{t} \\
& \text { and } \mathrm{i}_{\mathrm{c}}=\mathrm{e}_0 \omega \mathrm{C} \sin \left(\omega \mathrm{t}+\frac{\pi}{2}\right)=\mathrm{e}_0 \omega \mathrm{C} \cos \omega \mathrm{t}
\end{aligned}
\)
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