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AC Voltage Applied to a Series LCR Circuit - Phasor-diagram Solution

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Estimated time: 8 minutes
CBSE: Class 12

Introduction

A resistor R, inductor L, and capacitor C are connected in series with an AC (alternating current) source.

Key Property

Since all three elements are in series, the same AC current flows through each of them. This current has the same amplitude and same phase throughout the circuit.

The instantaneous current in the circuit is written as:

i = im sin⁡ (ωt + ϕ)   ...(1)

where:

  • im​ = peak (maximum) amplitude of current
  • ω = angular frequency of the AC source
  • ϕ = phase difference between the applied voltage and the current
CBSE: Class 12

Phasors and Voltage Relationships

The circuit is analysed using phasors — rotating vectors that represent sinusoidal quantities.

The phasors involved are: I (current), VR, VL, VC, and V (applied voltage).

Phase Relationships Between Voltage and Current

Component Voltage Phasor Phase Relation with Current
Resistor (R) \[V_R\] Voltage is in phase with the current (\[\phi = 0^\circ\]).
Capacitor (C) \[V_C\] Voltage lags the current by \[\dfrac{\pi}{2}\] (90°).
Inductor (L) \[V_L\] Voltage leads the current by \[\dfrac{\pi}{2}\] (90°).
CBSE: Class 12

Voltage Amplitudes

The peak (amplitude) of voltage across each element is:

VRm = imR   ...(2)
VCm = imXC   ...(3)
VLm = imX  ...(4)

where XC​ and XL​ are the capacitive reactance and inductive reactance, respectively.

CBSE: Class 12

Phasor Sum (KVL in Phasor Form)

By Kirchhoff's Voltage Law, the phasors of the three voltages must add up to the applied voltage phasor:

VL + VR + VC = V   ...(5)
This is a phasor (vector) sum, not an algebraic sum. The voltages are at different phases and cannot be added directly as numbers.
CBSE: Class 12

Derivation of Impedance and Phase Angle

Using the right-angle triangle formed in the phasor diagram:

From the right-angle phasor triangle:

\[v_m^2=V_{Rm}^2+(V_{Cm}-V_{Lm})^2=(i_mR)^2+(i_mX_C-i_mX_L)^2\]
\[v_m=i_m\sqrt{R^2+(X_C-X_L)^2}\]

Therefore, the peak current is:

\[{i_m=\frac{v_m}{Z}}\]   ​​​...(6)

where Impedance Z is defined as:

\[{Z=\sqrt{R^{2}+(X_{C}-X_{L})^{2}}}\]   ​​...(7)

Phase Angle φ

From the phasor triangle, the phase angle between the applied voltage and current is:

\[{\tan\phi=\frac{X_C-X_L}{R}}\]   ​​​...(8)
CBSE: Class 12

Phase Conditions

Depending on the relative magnitudes of XC​ and XL​, the circuit behaves differently.

Condition Circuit Nature Behaviour
\[X_C > X_L\] Predominantly Capacitive Current leads the voltage (\[\phi > 0)\].
\[X_C < X_L\] Predominantly Inductive Current lags the voltage (\[\phi < 0)\].
\[X_C = X_L\] Resonance Current and voltage are in phase (\[\phi = 0)\].
CBSE: Class 12

Steady-State vs. Transient Solution

The phasor method gives only the steady-state solution of the circuit — the long-term sinusoidal behaviour after the circuit has settled.

The complete solution is:

Complete Solution = Transient Response + Steady-State (Phasor) Solution

The transient response appears when the circuit is first switched on, but it dies out over time, leaving only the steady-state behaviour.

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