Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.
A balanced Y-connected load with an impedance of 4 + j3 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 50 kW at a power factor of 0.75 lagging from a 415 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 = 4:1 has 480 V applied to its primary winding. A 12 Ω resistor is connected across the secondary winding. Find the primary current I1.
A single-phase 500 kVA transformer has a core loss of 1500 W. At rated load and a power factor of 0.85 lagging its efficiency is 96%. Find the full-load copper loss Pcu,fl in watts.
Two loads are connected in parallel to a 400 V single-phase supply: an 8 kW load at a power factor of 0.65 lagging and a 9 kW load at a power factor of 0.8 lagging. Find the total current I drawn from the supply.
A single-phase load takes 10 kW at a power factor of 0.65 lagging from a 240 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.9. Find the reactive power QC the capacitor must supply.
A series RLC circuit is connected to a 24 V (rms), 500 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).
A series RLC circuit is connected to a 24 V (rms), 500 Hz supply. Find the rms current I.
A DC motor with k = 0.8 V·s/rad runs from 110 V. Neglecting all losses at no load, find its no-load speed n0 in rpm.
A separately excited DC motor (k = 0.8 V·s/rad at full field, Ra = 2 Ω, V = 240 V) drives a constant load torque of 6.4 N·m. The field flux is reduced to 85 % of its full value. Find the new speed n.
Answer: P = 27.56 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: Cph = 511.2 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 = 2.5 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 = 192 Ω, and 480/192 gives the same current.
Answer: Pcu,fl = 16208 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: I = 58.54 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: QC = 6.848 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.
The real power P does not change; only the reactive power drawn from the supply does.
Answer: φ = 77.92 °
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: I = 0.1281 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: n0 = 1313 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 = 3106 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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