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 18.5 kW of mechanical output power at full load. At this load its efficiency is 94% and its power factor is 0.84 lagging. The supply has a line-to-line voltage of 400 V. Find the line current IL.
A balanced three-phase load takes 75 kW at a power factor of 0.75 lagging from a 480 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.95 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 60 Ω resistor is connected across the secondary winding. Find the primary current I1.
A single-phase 150 kVA transformer has a core loss of 750 W and a copper loss of 1875 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 26 kW load at a power factor of 0.8 lagging and an 11 kW load at a power factor of 0.9 leading. Find the overall power factor cos φ seen by the supply.
A 50 kVA single-phase transformer supplies a load of 25 kW at a power factor of 0.85 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), 60 Hz supply. Find the rms current I.
A resistor R1 is connected in series with a parallel combination of R2 and a capacitor C, across a 24 V (rms), 500 Hz supply. Find the magnitude of the impedance |Z|.
A DC motor with k = 0.3 V·s/rad runs from 48 V. Neglecting all losses at no load, find its no-load speed n0 in rpm.
A separately excited DC motor on 48 V (Ra = 0.25 Ω) runs at 1200 rpm with an armature current of 20 A. The load changes and the current becomes 12 A, with the field unchanged. Find the new speed n2.
Answer: IL = 33.82 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 = 573.2 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 = 0.2 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 = 6000 Ω, and 1200/6000 gives the same current.
Answer: Sη,max = 94.87 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: cos φ = 0.9338
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: ΔP = 22.54 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: I = 0.2436 A
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: |Z| = 32.83 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|.
Multiply top and bottom by the complex conjugate to split Zp into real and imaginary parts.
Answer: n0 = 1528 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 = 1256 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.
ExamProblems · examproblems.com/courses/power-systems