Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.
A three-phase induction motor delivers 30 kW of mechanical output power at full load. At this load its efficiency is 95% and its power factor is 0.82 lagging. The supply has a line-to-line voltage of 400 V. Find the line current IL.
A balanced three-phase load takes 300 kW at a power factor of 0.8 lagging from a 480 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.98 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.
A single-phase 250 kVA, 7200/240 V transformer operates at its rated load. Treat it as ideal. Find the current IHV in the high-voltage winding.
A single-phase 500 kVA transformer has a core loss of 3000 W and a copper loss of 6000 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?
Two loads are connected in parallel to a 240 V single-phase supply: a 14 kW load at a power factor of 0.6 lagging and a 5 kW load at a power factor of 0.8 lagging. Find the total current I drawn from the supply.
A 12 V (rms), 1000 Hz source has an internal impedance Zs = 50 + j5 Ω. The load impedance ZL can be chosen freely. The load for maximum power is built from a resistor in series with a capacitor. Find the capacitance C.
A resistor and an inductor are connected in parallel across a 50 V (rms), 60 Hz supply. Find the magnitude of the impedance |Z|.
A series RLC circuit is connected to a 24 V (rms), 500 Hz supply. Find the rms voltage VC across the capacitor.
A DC motor (k = 0.1 N·m/A, Ra = 0.25 Ω) runs from 24 V and drives a load torque of 0.4 N·m. Find its speed n in rpm.
A separately excited DC motor (k = 0.5 V·s/rad at full field, Ra = 0.2 Ω, V = 110 V) drives a constant load torque of 12.5 N·m. The field flux is reduced to 75 % of its full value. Find the new speed n.
Answer: IL = 55.59 A
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.
Answer: Cph = 1889 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: IHV = 34.72 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.
Both windings carry the same apparent power.
Answer: Sη,max = 353.6 kVA
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.
Answer: I = 122.4 A
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.
A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).
Answer: C = 31.83 µF
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.
The load reactance cancels the source reactance, and the load resistance equals the source resistance.
Answer: |Z| = 17.98 Ω
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|.
Both branches are connected straight across the supply, so each sees the full voltage V.
The resistor current is in phase with V and the reactive current is 90° out of phase, so they add like the sides of a right triangle.
Answer: VC = 10.51 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|.
In series the impedances add as complex numbers: Z = R + j(XL − XC).
The same current flows through every element of a series circuit.
Answer: n = 2196 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.
Answer: n = 2631 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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