In a series A.C. circuit,the current and volt | vedclass

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(C) Given:$I = 10 \sin \omega t$$V = 10 \sin \left( \omega t + \pi / 4 \right)$Comparing these with the general forms $I = I_0 \sin \omega t$ and $V =

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$(b), (d)$

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(C) Given:$I = 10 \sin \omega t$$V = 10 \sin \left( \omega t + \pi / 4 \right)$Comparing these with the general forms $I = I_0 \sin \omega t$ and $V = V_0 \sin (\omega t + \phi)$,we see that the phase difference is $\phi = \pi / 4$.Since the voltage leads the current by a phase angle of $\pi / 4$ (positive phase angle),the circuit must be inductive in nature.An $RL$ circuit (Resistor and Inductor) results in a voltage that leads the current.An $LCR$ circuit can also result in a voltage that leads the current if the inductive reactance $X_L$ is greater than the capacitive reactance $X_C$ (i.e.,$X_L > X_C$).Therefore,the circuit could contain both $L$ and $R$,or it could contain all three elements $L, C,$ and $R$ provided the net reactance is inductive.Thus,options $(b)$ and $(d)$ are correct.

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In a series $A.C.$ circuit,the current and voltage are given respectively as:$I = 10 \sin \omega t$; $V = 10 \sin \left( \omega t + \pi / 4 \right)$Then,which of the following elements may be present in the circuit?$(a)$ Only $C$$(b)$ All $L, C$ and $R$$(c)$ Both $C$ and $R$$(d)$ Both $L$ and $R$