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A series LCR circuit is connected to an ac source. Using the phasor diagram, derive the expression for the impedance of the circuit.

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Question

A series LCR circuit is connected to an ac source. Using the phasor diagram, derive the expression for the impedance of the circuit.

Derivation
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Solution


LCR Circuit

E = E0 sinωt is applied to a series LCR circuit. Since all three of them are connected in series the current through them is the same. But the voltage across each element has a different phase relation with the current.

The potential difference VL, VC and VR across L, C and R at any instant is given by VL = IXL, VC = IXC and VR = IR, where I is the current at that instant.

VR is in phase with I. VL leads I by 90° and VC lags behind I by 90° so the phasor diagram will be as shown Assuming VL > VC, the applied emf E which is equal to the resultant of the potential drop across R, L & C is given as

`E^2 = I^2[R^2 + (X_L - X_C)^2]`

Or I = `E/sqrt[[R^2 + (X_L - X_C)^2]` = `E/Z`, Where Z is Impedance.

Emf leads current by a phase angle φ as `tanφ = (V_L - V_C)/R = (X_L - X_C)/R`

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2023-2024 (March) Board Sample Paper

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