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 37 kW of mechanical output power at full load. At this load its efficiency is 92% and its power factor is 0.84 lagging. The supply has a line-to-line voltage of 690 V. Find the line current IL.
A balanced three-phase load takes 60 kW at a power factor of 0.78 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 0.92 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.
An ideal transformer with a turns ratio N₁:N₂ = 4:1 has 920 V applied to its primary winding. A 15 Ω resistor is connected across the secondary winding. Find the primary current I1.
A single-phase 25 kVA transformer has a core loss of 75 W. At rated load and a power factor of 0.8 lagging its efficiency is 97.5%. Find the full-load copper loss Pcu,fl in watts.
A single-phase motor takes 24 kW at a power factor of 0.75 lagging from a 230 V supply. Find the reactive power Q it draws.
A 150 kVA single-phase transformer supplies a load of 99 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 a capacitor are connected in parallel across a 50 V (rms), 200 Hz supply. Find the magnitude of the impedance |Z|.
A series RLC circuit is driven by a 2 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 with k = 0.12 V·s/rad runs from 24 V. Neglecting all losses at no load, find its no-load speed n0 in rpm.
A separately excited DC motor on 240 V (Ra = 1.5 Ω) runs at 1800 rpm with an armature current of 6 A. The load changes and the current becomes 12 A, with the field unchanged. Find the new speed n2.
Answer: IL = 40.06 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: QC = 22.58 kvar
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.
Answer: I1 = 3.833 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 = 240 Ω, and 920/240 gives the same current.
Answer: Pcu,fl = 437.8 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.
Answer: Q = 21.17 kvar
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 = 31.33 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| = 7.03 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: VL = 20.63 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 = 10.31 times the supply voltage.
Answer: n0 = 1910 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: n2 = 1730 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.
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