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 20 + j15 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 415 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 0.75 lagging from a 690 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 = 10:1 has 2400 V applied to its primary winding. A 15 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 6400 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.
A single-phase motor takes 3 kW at a power factor of 0.85 lagging from a 240 V supply. Find the line current I.
A 200 kVA single-phase transformer supplies a load of 84 kW at a power factor of 0.65 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 10 V (rms), 100 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).
A resistor R1 is connected in series with a parallel combination of R2 and a capacitor C, across a 100 V (rms), 50 Hz supply. Find the magnitude of the impedance |Z|.
A DC motor (k = 1.5 N·m/A, Ra = 1 Ω) runs from 110 V and drives a load torque of 12 N·m. Find its speed n in rpm.
A DC motor (k = 0.25 V·s/rad, Ra = 2 Ω) on a fixed 48 V supply drives a load of 0.5 N·m. Find the series resistance Rs that slows it down to 900 rpm.
Answer: P = 16.53 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| = 0.8, and √3·VL·IL·cos φ gives the same power.
Answer: QC = 91.18 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 = 1.6 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 = 1500 Ω, and 2400/1500 gives the same current.
Answer: Ploss = 5200 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 = 96.29%.
Answer: I = 14.71 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 = 90.23 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 = −71.87 °
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: |Z| = 106.9 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: n = 649.4 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: Rs = 10.22 Ω
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.
The resistor wastes 40.88 W; a chopper controls the speed with far less loss.
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