Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.
A balanced three-phase load draws a line current of 25 A from a supply with a line-to-line voltage of 208 V. The power factor is 0.85 lagging. Find the total real power P drawn by the load in kW.
A balanced three-phase load takes 200 kW at a power factor of 0.65 lagging from a 690 V (line-to-line), 50 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 = 10:1 has 1200 V applied to its primary winding. A 4 Ω resistor is connected across the secondary winding. Find the primary current I1.
A single-phase 200 kVA transformer has a core loss of 600 W and a copper loss of 1800 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?
A single-phase motor takes 28 kW at a power factor of 0.6 lagging from a 400 V supply. Find the line current I.
A 100 kVA single-phase transformer supplies a load of 29 kW at a power factor of 0.8 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 resistor and an inductor are connected in parallel across a 12 V (rms), 200 Hz supply. Find the magnitude of the impedance |Z|.
A series RLC circuit is connected to a 230 V (rms), 50 Hz supply. Find the rms voltage VC across the capacitor.
A DC motor draws 20 A from a 120 V supply. Its armature resistance is 0.8 Ω and its rotational (friction, windage and iron) losses are 104 W. Find its efficiency η.
A DC motor with k = 1.2 V·s/rad and Ra = 0.4 Ω must drive a load of 18 N·m at 500 rpm. Find the armature voltage V it needs.
Answer: P = 7.656 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.
Answer: Cph = 915.7 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 = 3 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 = 400 Ω, and 1200/400 gives the same current.
Answer: Sη,max = 115.5 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 = 116.7 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.
Only the real part of the power does work, so divide by the power factor.
Answer: ΔP = 68.61 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: |Z| = 14.59 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 = 240.8 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: η = 82.33 %
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: V = 68.83 V
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.
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