In the AC circuit shown,E = E_0 sin(omega t + | vedclass

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(C) The phase difference between voltage and current is given by $\Delta \phi = \phi_i - \phi_e = (\omega t + \phi + \frac{\pi}{4}) - (\omega t + \phi

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$C$ and $R$ in series or $L, C$ and $R$ in series

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(C) The phase difference between voltage and current is given by $\Delta \phi = \phi_i - \phi_e = (\omega t + \phi + \frac{\pi}{4}) - (\omega t + \phi) = +\frac{\pi}{4}$.Since the phase angle is positive,the current leads the voltage,which indicates that the circuit is capacitive in nature.In an $AC$ circuit,if the current leads the voltage,the net reactance must be capacitive $(X_C > X_L)$.This condition is satisfied if the circuit contains a capacitor and a resistor ($C-R$ circuit) or a combination of an inductor,a capacitor,and a resistor ($L-C-R$ circuit) where the capacitive reactance dominates the inductive reactance.

In the $AC$ circuit shown,$E = E_0 \sin(\omega t + \phi)$ and $i = i_0 \sin(\omega t + \phi + \frac{\pi}{4})$. Then,the box contains:

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