Resonance, Coupled Coils and Three-Phase
Series and parallel resonance, quality factor and bandwidth, mutual inductance and dot convention, balanced star and delta systems, line and phase quantities, and the two-wattmeter method.
This lesson covers three AC topics that GATE EE tests every year: resonance, magnetically coupled coils, and three-phase circuits.
Series resonance
At ω₀ = 1/√(LC), X_L = X_C, so Z = R (minimum) and the current is maximum, in phase with the voltage.
- Quality factor Q = ω₀L/R = 1/(ω₀CR) = (1/R)√(L/C).
- Bandwidth BW = ω₀/Q = R/L (rad/s). Half-power frequencies: ω₁ω₂ = ω₀².
- Voltage across L or C at resonance = Q × source voltage (voltage magnification).
Parallel resonance
An ideal parallel LC tank has maximum impedance (infinite) at ω₀; with a coil of resistance R, ω₀ = √(1/(LC) − R²/L²) and the dynamic impedance is L/(CR). Current magnification instead of voltage.
Coupled coils
- Mutual inductance M = k√(L₁L₂), with coupling coefficient 0 ≤ k ≤ 1.
- Series aiding: L = L₁ + L₂ + 2M; series opposing: L₁ + L₂ − 2M.
- Dot convention: current entering both dots gives aiding fluxes (+M).
- Ideal transformer: V₁/V₂ = N₁/N₂, and an impedance Z_L on the secondary looks like (N₁/N₂)² Z_L from the primary.
Balanced three-phase systems
| Connection | Line voltage | Line current |
|---|---|---|
| Star (Y) | V_L = √3 V_ph (leads by 30°) | I_L = I_ph |
| Delta (Δ) | V_L = V_ph | I_L = √3 I_ph (lags by 30°) |
- Total power P = √3 V_L I_L cos φ (φ = phase angle of the load impedance).
- A balanced delta load Z_Δ is equivalent to a star load Z_Δ/3.
- Phase sequence ABC: V_B lags V_A by 120°.
Two-wattmeter method (three-wire balanced load):
- P = W₁ + W₂.
- tan φ = √3 (W₁ − W₂)/(W₁ + W₂).
- At pf 0.5, one wattmeter reads zero; below 0.5, one reads negative.
Worked example
A balanced star load of (8 + j6) Ω per phase on a 400 V line supply.
- V_ph = 400/√3 = 230.9 V; |Z| = 10 Ω; I = 23.09 A; pf = 0.8.
- P = √3 × 400 × 23.09 × 0.8 = 12.8 kW.
Practice
A series RLC circuit has R = 10 Ω, L = 0.1 H and C = 10 µF. What is its quality factor?
Show answer and explanation
Answer: 10Q = (1/R)√(L/C) = 0.1 × √10,000 = 10.
In the two-wattmeter method, W₁ = 3 kW and W₂ = 1 kW. What is the power factor? (Give 3 decimal places.)
Show answer and explanation
Answer: 0.756tan φ = √3 × 2/4 = 0.866, so φ = 40.9° and cos φ = 0.756.
Two coils of 4 mH and 9 mH have k = 0.5. What is the total inductance (mH) in series aiding?
Show answer and explanation
Answer: 19M = 0.5√36 = 3 mH; L = 4 + 9 + 6 = 19 mH.
In the worked example, what is the total power in kW? (Give 1 decimal place.)
Show answer and explanation
Answer: 12.8P = 3I²R = 3 × 23.09² × 8 = 12,800 W = 12.8 kW.