Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.
A balanced Δ-connected load with an impedance of 15 + j20 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 208 V. Find the total real power P absorbed by the load in kW.
A balanced three-phase load takes 200 kW at a power factor of 0.65 lagging from a 400 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.9 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.
An ideal transformer with a turns ratio N₁:N₂ = 4:1 has 920 V applied to its primary winding. A 5 Ω resistor is connected across the secondary winding. Find the primary current I1.
A single-phase 200 kVA transformer has a core loss of 1600 W and a copper loss of 4800 W at rated load. It supplies 75% of its rated kVA at a power factor of 0.9 lagging. Find the total losses Ploss in watts at this load.
An inductive load takes 14.5 A and 1218 W from a 120 V (rms) single-phase supply. Model it as a resistance R in series with a reactance X. Find R.
A 200 kVA single-phase transformer supplies a load of 66 kW at a power factor of 0.6 lagging. How much extra real power ΔP can be added as a resistive (unity power factor) load before the transformer reaches its kVA rating?
A series RLC circuit is connected to a 120 V (rms), 1000 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).
A series RLC circuit is driven by a 12 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.
A DC motor (k = 0.2 V·s/rad, Ra = 1 Ω) runs from 24 V. At full load it draws 2 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.
A separately excited DC motor (k = 0.75 V·s/rad at full field, Ra = 0.4 Ω, V = 110 V) drives a constant load torque of 22.5 N·m. The field flux is reduced to 85 % of its full value. Find the new speed n.
Answer: P = 3.115 kW
Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ.
Only the resistive part of each impedance absorbs real power.
Check: cos φ = R/|Z| = 0.6, and √3·VL·IL·cos φ gives the same power.
Answer: Cph = 2271 µF
Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ. Power factor correction: capacitors supply reactive power locally, QC = P·(tan φ1 − tan φ2). Each of the three capacitors supplies QC/3 = V²·ω·C, where V is the voltage across that capacitor: VL in a Δ bank, VL/√3 in a Y bank.
The real power is unchanged, so each power factor fixes the reactive power: Q = P·tan φ.
In a Y bank each capacitor is connected between a line and the star point.
Answer: I1 = 11.5 A
Ideal transformer: the voltages are in the ratio of the turns and the currents in the inverse ratio: V₁/V₂ = N₁/N₂ = a and I₁/I₂ = N₂/N₁ = 1/a, so the power in equals the power out, V₁·I₁ = V₂·I₂. A load impedance Z on the secondary looks like a²·Z from the primary.
An ideal transformer has no losses: V₁·I₁ = V₂·I₂.
Check: the load seen from the primary is a²·RL = 80 Ω, and 920/80 gives the same current.
Answer: Ploss = 4300 W
Transformer losses: the core loss (hysteresis and eddy currents) depends on the voltage and is the same at any load. The copper loss in the windings goes with the current squared, so at a fraction x of rated load it is x²·Pcu,fl. Efficiency η = Pout/(Pout + Pcore + Pcu), with Pout = x·S·pf. The efficiency is highest at the load where the copper loss equals the core loss.
The core loss is the same at any load.
The efficiency at this load is η = 96.91%.
Answer: R = 5.793 Ω
AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.
Answer: ΔP = 113.6 kW
AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.
At full rating S = Smax, and Q is unchanged by the resistive load.
Answer: φ = 71.69 °
AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.
In series the impedances add as complex numbers: Z = R + j(XL − XC).
Answer: VL = 72.36 V
AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.
At resonance the reactances cancel, so the impedance is just R.
This is Q = 6.03 times the supply voltage.
Answer: SR = 9.091 %
DC motor: the armature is a resistance Ra in series with the back-EMF E, so V = E + Ia·Ra. The back-EMF grows with speed, E = k·ω, and the torque with current, T = k·Ia, with the same motor constant k (V·s/rad = N·m/A) in SI units. Convert speeds with ω = 2π·n/60. The developed power is E·Ia = T·ω; the input power is V·Ia, and the copper loss Ia²Ra and the rotational loss make the difference. The speed falls in a straight line with torque, ω = V/k − Ra·T/k², from the no-load speed V/k to zero at the stall torque k·V/Ra. At standstill E = 0, so the starting current V/Ra is large unless a resistor limits it.
Answer: n = 1436 rpm
DC motor: the armature is a resistance Ra in series with the back-EMF E, so V = E + Ia·Ra. The back-EMF grows with speed, E = k·ω, and the torque with current, T = k·Ia, with the same motor constant k (V·s/rad = N·m/A) in SI units. Convert speeds with ω = 2π·n/60. The developed power is E·Ia = T·ω; the input power is V·Ia, and the copper loss Ia²Ra and the rotational loss make the difference. The speed falls in a straight line with torque, ω = V/k − Ra·T/k², from the no-load speed V/k to zero at the stall torque k·V/Ra. At standstill E = 0, so the starting current V/Ra is large unless a resistor limits it.
Field weakening is how a DC drive runs above its base speed.
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