In an LCR series circuit,if V is the effectiv | vedclass

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(C) In an $LCR$ series circuit,the current $I$ is the same through all components. The voltage across the resistor $V_R$ is in phase with the current.

Vedclass · Accepted Answer

$V^2=V_{R}^2+\left(V_{L}-V_{C}\right)^2$

Vedclass · Answer

(C) In an $LCR$ series circuit,the current $I$ is the same through all components. The voltage across the resistor $V_R$ is in phase with the current. The voltage across the inductor $V_L$ leads the current by $90^\circ$,and the voltage across the capacitor $V_C$ lags behind the current by $90^\circ$.Thus,$V_L$ and $V_C$ are in opposite directions. The net reactive voltage is $(V_L - V_C)$.Using the phasor diagram,the resultant voltage $V$ is the vector sum of $V_R$ and $(V_L - V_C)$,which are perpendicular to each other.Applying the Pythagorean theorem: $V^2 = V_R^2 + (V_L - V_C)^2$.

In an $LCR$ series circuit,if $V$ is the effective value of the applied voltage,$V_R$ is the voltage across $R$,and $V_L$ and $V_C$ are the effective voltages across $L$ and $C$ respectively,then:

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