Resonance in a series LCR circuit
At resonance X_L = X_C, so Z = R and the current is greatest. The resonant angular frequency is ω₀ = 1/√(LC).
What happens at resonance
In a series LCR circuit, X_L = ωL rises with frequency while X_C = 1/(ωC) falls. At one frequency, ω₀, the two become equal. This is resonance. Then V_L and V_C are equal and opposite, so they cancel. The impedance falls to its least value, Z = R, and the current is greatest, iₘ = vₘ/R. The current is in phase with the voltage (φ = 0). At resonance, V_L and V_C can each be much larger than the source voltage. Resonance needs both L and C; an RL or RC circuit cannot resonate.
Finding the resonant frequency
- 1The peak current is iₘ = vₘ/√(R² + (X_C − X_L)²).
- 2For fixed vₘ and R, iₘ is greatest when X_C − X_L = 0.
- 3So X_L = X_C, that is ω₀L = 1/(ω₀C).
- 4Rearranging: ω₀² = 1/(LC).
- 5★So, ω₀ = 1/√(LC), and the resonant frequency is ν₀ = 1/(2π√(LC)).
★ key step
How to draw the current to frequency graph
- Draw the horizontal axis for angular frequency ω and the vertical axis for peak current iₘ.
- Mark ω₀ on the horizontal axis.
- Draw a curve that rises from near zero, reaches a peak at ω₀, and falls again on the other side.
- Draw a second curve, for a larger resistance, with its peak at the same ω₀ but lower.
- Label the taller curve 'small R' and the lower curve 'large R'.
Worked example: resonance
A series LCR circuit has L = 2 mH, C = 5 μF and R = 10 Ω. It is connected to a source of peak voltage 20 V whose frequency can be varied. Find (a) the resonant angular frequency, (b) the peak current at resonance and (c) the peak voltage across the inductor at resonance.
- (a) LC = 2 × 10⁻³ × 5 × 10⁻⁶ = 10 × 10⁻⁹ = 10⁻⁸ s².
- ω₀ = 1/√(LC) = 1/√(10⁻⁸) = 1/10⁻⁴ = 10⁴ rad/s.
- (b) At resonance Z = R, so iₘ = vₘ/R = 20 V ÷ 10 Ω = 2 A.
- (c) X_L = ω₀L = 10⁴ × 2 × 10⁻³ = 20 Ω, so the peak voltage across L = iₘX_L = 2 A × 20 Ω = 40 V.
- This is twice the source voltage. The voltage across C is also 40 V and cancels it.
(a) 10⁴ rad/s (ν₀ = 10⁴/2π ≈ 1590 Hz); (b) 2 A; (c) 40 V
Use in tuning a radio
The antenna of a radio picks up signals from many stations at once. Its tuning circuit is an LCR circuit, usually with a variable capacitor. Turning the knob changes C, and so changes ω₀. When ω₀ matches the frequency of one station, the current at that frequency is largest. That station is heard clearly, while the others are too weak to be heard. This is tuning.
When you turn the tuning knob of an FM radio to pick up Akashvani (All India Radio), you change a capacitor until the circuit's resonant frequency matches that station's frequency.
2 marks: “State the condition for resonance in a series LCR circuit and write the expression for the resonant frequency.” X_L = X_C; ω₀ = 1/√(LC), so ν₀ = 1/(2π√(LC)).
3 marks: “Draw graphs showing how the current in a series LCR circuit varies with frequency for two values of resistance. Explain their shape and give one use of resonance.” Both curves peak at the same ω₀; the smaller R gives a taller peak since iₘ = vₘ/R at resonance; used in tuning radio and TV sets.
1 marks: “What is the phase difference between the current and the voltage in a series LCR circuit at resonance?” Zero; they are in phase.
Resonance: X_L = X_C, Z = R, φ = 0, and ω₀ = 1/√(LC), 'one over root LC'.