What is the impedance of an L-C-R series AC c | vedclass

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Question

(A) In an $L-C-R$ series $AC$ circuit,the voltage across the resistor $(V_R)$ is in phase with the current $(I)$.The voltage across the inductor $(V_L

Vedclass · Accepted Answer

$Z = \sqrt{R^2 + (X_L - X_C)^2}$

Vedclass · Answer

(A) In an $L-C-R$ series $AC$ circuit,the voltage across the resistor $(V_R)$ is in phase with the current $(I)$.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 $90^\circ$.The net reactance of the circuit is $(X_L - X_C)$.Using the phasor diagram,the total voltage $V$ is given by $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.Since $V = IZ$,$V_R = IR$,$V_L = IX_L$,and $V_C = IX_C$,we substitute these into the equation:$IZ = \sqrt{(IR)^2 + (IX_L - IX_C)^2}$$IZ = I \sqrt{R^2 + (X_L - X_C)^2}$Therefore,the impedance $Z$ is $Z = \sqrt{R^2 + (X_L - X_C)^2}$.

What is the impedance of an $L-C-R$ series $AC$ circuit?

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