The series LCR circuit and resonance

Adding voltages with phasors

Narrated video. Transcript:

Now put a resistor R, an inductor L and a capacitor C in series with an ac source, v equals v m sin ωt. Because they are in series, the same current i flows through all three at every instant.

So start the phasor diagram with the current phasor I. Let the current be i equals i m sin of ωt plus φ, where φ is the phase difference we want to find.

The voltage across the resistor, V R, is in phase with I. Its length is i m R.

The voltage across the inductor, V L, leads the current by 90°. Its length is i m X L. The voltage across the capacitor, V C, lags the current by 90°. Its length is i m X C.

So V L and V C point in exactly opposite directions. They partly cancel. Their resultant has length i m times X C minus X L.

The source voltage phasor V is the vector sum of V R, V L and V C. V R and the net reactive part are at right angles, so by Pythagoras, v m² equals i m R squared plus i m times X C minus X L, squared.

That gives i m equals v m over the square root of R² plus X C minus X L squared. The denominator is the impedance Z. So i m equals v m over Z.

Draw R along the x-axis and X C minus X L upward. Z is the hypotenuse of this impedance triangle, and its angle is φ. So tan φ equals X C minus X L over R.

If X C is bigger than X L, φ is positive: the circuit is mainly capacitive and the current leads. If X L is bigger, φ is negative: the circuit is inductive and the current lags.

A worked example

Narrated video. Transcript:

A resistor of 200 ohms and a capacitor of 15 microfarads are connected in series to a 220 volt, 50 hertz ac source. There is no inductor, so X L is zero.

First the reactance. X C equals 1 over 2π × 50 × 15 × 10⁻⁶, which is 212.2 ohms.

The impedance is Z equals √(200² plus 212.2²), which is √85 029, about 291.6 ohms.

So the rms current is I equals V over Z, 220 over 291.6, which is about 0.754 amperes.

The voltage across the resistor is V R equals I R, 0.754 × 200, about 151 volts. Across the capacitor, V C equals I X C, 0.754 × 212.2, about 160 volts.

Wait. 151 plus 160 is 311 volts, far more than the 220 volts of the source! Has something gone wrong?

No. The two voltages are 90° out of phase, so they never reach their peaks at the same moment. They add like phasors, at right angles.

√(151² plus 160²) is about 220 volts: just the source voltage.

And the phase: tan φ equals X C over R, 212.2 over 200, about 1.06. So φ is about 46.7°, and the current leads the voltage, as it should in a capacitive circuit.

Resonance

Narrated video. Transcript:

Keep v m fixed and slowly turn up the frequency of the source. X L equals ωL rises. X C equals 1 over ωC falls. At one frequency they become equal.

There, X C minus X L is zero, so Z equals R, its smallest possible value. The current amplitude is largest: i m equals v m over R. This is resonance.

Setting ω₀L equal to 1 over ω₀C gives the resonant angular frequency, ω₀ equals 1 over √(LC). At resonance the current is in phase with the voltage.

Plot i m against ω for L equals 1 millihenry, C equals 1 nanofarad and v m equals 100 volts. ω₀ is 1 over √(10⁻³ × 10⁻⁹), which is 10⁶ radians per second.

With R equals 100 ohms the peak is 1 ampere, and tall and narrow. With R equals 200 ohms the peak is only 0.5 amperes, and much flatter.

How sharp is the peak? Mark the two frequencies where the current has fallen to i m max over √2. They are ω₀ plus Δω and ω₀ minus Δω. The gap between them, 2Δω, is the bandwidth, and Δω equals R over 2L.

The quality factor Q equals ω₀ over 2Δω, which is ω₀L over R. For R equals 100 ohms, Q equals 10⁶ × 10⁻³ over 100, which is 10. A larger Q means a sharper resonance.

A radio uses this. The antenna picks up many stations at once. Turning the tuning knob changes C, until ω₀ matches one station. Only that station drives a large current.

Example: L equals 5 henry, C equals 80 microfarad, R equals 40 ohms, on a 230 volt variable-frequency source. ω₀ equals 1 over √(5 × 80 × 10⁻⁶), which is 50 radians per second.

At resonance Z equals 40 ohms, so I equals 230 over 40, which is 5.75 amperes. X L equals 250 ohms, so V L and V C are each 5.75 × 250, about 1438 volts. They cancel exactly, and V R is the full 230 volts.

Try it: phasors in series LCR

Interactive simulation. Things to try:

• At 5 Hz, X_C > X_L: the circuit is capacitive, φ is positive and the I phasor is ahead of V.
• Raise the frequency above 8 Hz. X_L wins, φ turns negative and the current lags.
• Press “Go to resonance”. V_L and V_C (orange and purple) cancel exactly, Z = R and the current is in phase with V.
• At resonance read V_L and V_C: about 1440 V each, from a 230 V source! Only their difference matters.

Try it: the resonance curve

Interactive simulation. Things to try:

• Drag across the graph to change the source frequency ω. The current peaks at ω₀ = 1/√(LC) = 10⁶ rad/s.
• The dashed curve is R = 200 Ω. Halving R doubles the peak and halves the bandwidth 2Δω = R/L.
• Read Q = ω₀L/R. Make R smaller: Q rises and the curve gets sharper.
• Change C, as a radio tuning knob does: the whole peak slides to a new ω₀.

Impedance

A series LCR circuit has R = 30 Ω, X_L = 80 Ω and X_C = 40 Ω, and is connected to a 220 V ac source. What is the rms current?

1. 1.47 A
2. 2.33 A
3. 3.14 A
4. 4.4 A

Resonance and Q

A series LCR circuit has L = 2 H, C = 50 μF and R = 10 Ω. What are its resonant angular frequency ω₀ and its quality factor Q?

1. ω₀ = 100 rad/s, Q = 20
2. ω₀ = 15.9 rad/s, Q = 20
3. ω₀ = 100 rad/s, Q = 0.05
4. ω₀ = 10⁴ rad/s, Q = 2000

Summary

• Impedance: Z = √(R² + (X_C − X_L)²), i_m = v_m/Z
• Phase of the current: tan φ = (X_C − X_L)/R
• Resonance: ω₀ = 1/√(LC), Z = R
• Sharpness: 2Δω = R/L, Q = ω₀L/R

• In series, the current is common. V_R is in phase with I, V_L leads it by 90° and V_C lags it by 90°; the voltages add as phasors, not as numbers.
• X_C > X_L: capacitive, current leads. X_L > X_C: inductive, current lags.
• At resonance X_L = X_C, the impedance is just R, the current is largest and in phase with the voltage. Resonance needs both L and C.
• A small R gives a tall, narrow peak (large Q). Radio tuning uses this.