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Question:
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In the $A C$ circuit shown,
$$
\begin{aligned}
& E=E_0 \sin (\omega t+\phi) \text { and } \\
& i=i_0 \sin \left(\omega t+\phi+\frac{\pi}{4}\right) .
\end{aligned}
$$
Then, the box contains
Options:
$$
\begin{aligned}
& E=E_0 \sin (\omega t+\phi) \text { and } \\
& i=i_0 \sin \left(\omega t+\phi+\frac{\pi}{4}\right) .
\end{aligned}
$$
Then, the box contains
Solution:
2622 Upvotes
Verified Answer
The correct answer is:
C and R in series or L, C and R in series
As current leads by $\frac{\pi}{4}$, so circuit must be more capacitive them inductive. Hence, it is either a $C-R$ combination or $L-C-R$ combination.
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