In the $A C$ circuit shown, $ \begin{aligned} & E=E_0 \sin (\omega t+\phi) \text { and } \\ & i=i_0 \sin…
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
Only C
L and R in series
C and R in series or L, C and R in series
Only R
Solution
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.
