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AC Voltage Applied to a Series LCR Circuit - Resonance

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Estimated time: 8 minutes
CBSE: Class 12

Definition: Resonance

Resonance in a series LCR circuit is the phenomenon that occurs at a particular frequency at which the amplitude of current is maximum. This happens when the inductive reactance equals the capacitive reactance: XC = XL.

CBSE: Class 12

Derivation of Resonant Frequency

Step 1: Current amplitude in series LCR circuit:

  • im = \[\frac {v_m}{Z}\]

where impedance:

  • Z = \[\sqrt {R^2+(X_C−X_L)^2}\]

Step 2: Condition for maximum current:

For im​ to be maximum, Z must be minimum.
Z is minimum when:

  • (XC − XL) = 0 ⇒ XC = XL

Step 3: Minimum impedance at resonance:

  • Zmin = \[\sqrt {R^2+0}\] = R

Step 4: Finding resonant frequency:

  • XC = XL
  • \[\frac {1}{ω_0C}\] = \[ω_0L\]
  • \[ω_0^2\] = \[\frac {1}{LC}\]
  • ω0 = \[\frac {1}{\sqrt {LC}}\]

Step 5: Maximum current at resonance:

  • \[{i_{max} = \frac{v_m}{R}}\]
CBSE: Class 12

Key Results at Resonance

Quantity At Resonance Reason
Condition \[X_C = X_L\] Inductive and capacitive reactances are equal and cancel each other.
Impedance Z = R (minimum) Since (\[X_C = X_L\]) = 0, the impedance is reduced to the resistance alone.
Current \[i_m = \dfrac{v_m}{R}\] (maximum) The impedance is minimum; therefore, the circuit current is maximum.
Nature of Circuit Purely resistive There is no net reactance, so voltage and current are in phase.
CBSE: Class 12

Example

Given: RC series circuit with:

  • R = 200 Ω
  • C = 15.0 μF = 15.0 × 10−6 F
  • AC source: 220 V, 50 Hz

Find: (a) Current in the circuit, (b) Voltage across R, (c) Voltage across C

Step 1: Angular frequency:

  • ω = 2πf = 2π × 50 = 314.16 rad/s

Step 2: Capacitive reactance:

  • XC = \[\frac{1}{\omega C}=\frac{1}{314.16\times15\times10^{-6}}\] ≈ 212.2Ω

Step 3: Impedance (RC circuit, no inductor):

  • Z = \[\sqrt{R^2+X_C^2}=\sqrt{(200)^2+(212.2)^2}=\sqrt{40000+45029}\]​ ≈ 291.7Ω

Step 4: RMS Current:

  • I = \[\frac{V_{rms}}{Z}=\frac{220}{291.7}\] ​≈ 0.755A

Step 5: Voltage across R:

  • VR = IR = 0.755 × 200 = 151 V

Step 6: Voltage across C:

  • VC = IXC = 0.755 × 212.2 ≈ 160.3 V

Step 7: Verification using phasor addition:

  • V = \[\sqrt{V_R^2+V_C^2}=\sqrt{(151)^2+(160.3)^2}=\sqrt{22801+25696}\]​ ≈ 220V
CBSE: Class 12

Real-Life Analogy - The Swing

Think of a child being pushed on a swing.

Swing (Mechanical) LCR Circuit (Electrical)
The swing has a natural frequency of oscillation. The LCR circuit has a natural resonant frequency \[\omega_0\].
Pushing the swing at its natural frequency produces the maximum amplitude of oscillation. Driving the LCR circuit at \[\omega_0\] produces the maximum current (resonance).
Friction limits the maximum amplitude of the swing. Resistance (R) limits the maximum current in the circuit.

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