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(B) In a series $LCR$ circuit,the current $I$ is in phase with the resistor voltage $V_R$. The net voltage $V$ is the vector sum of $V_R$,$V_L$,and $V

Vedclass · Accepted Answer

$I = I_0 \cos(\omega t' - \pi/6)$

Vedclass · Answer

(B) In a series $LCR$ circuit,the current $I$ is in phase with the resistor voltage $V_R$. The net voltage $V$ is the vector sum of $V_R$,$V_L$,and $V_C$. Let the current be along the $x$-axis. Then $V_R$ is along the $x$-axis,$V_L$ is along the positive $y$-axis,and $V_C$ is along the negative $y$-axis. The net reactive voltage is $V_L - V_C = 3 	ext{ V} - 2 	ext{ V} = 1 	ext{ V}$ (along the positive $y$-axis). The resistive voltage is $V_R = \sqrt{3} 	ext{ V}$ (along the $x$-axis). The phase angle $\phi$ between the source voltage $V$ and the current $I$ is given by $	an \phi = \frac{V_L - V_C}{V_R} = \frac{1}{\sqrt{3}}$. Thus,$\phi = 30^\circ = \pi/6$. Since the net voltage leads the current by $\phi = \pi/6$,the current lags behind the voltage by $\pi/6$. If $V = V_0 \cos(\omega t')$,then $I = I_0 \cos(\omega t' - \pi/6)$.

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