Power Systems and Electrical Machines: cram pack

Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.

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1Formulas and methods

Three-phase power

Y connection
VL = √3·Vph, IL = Iph
Δ connection
VL = Vph, IL = √3·Iph
Total real power
P = √3·VL·IL·cos φ = 3·Iph²·R
Apparent and reactive power
S = √3·VL·IL, Q = P·tan φ
Correction capacitors
QC = P(tan φ1 − tan φ2), C = QC/(3ωV²)
Voltage across each capacitor
Δ: VL Y: VL/√3
Motor input
Pin = Pout/η

Transformers

Turns ratio
a = N1/N2 = V1/V2
Current ratio
I1/I2 = N2/N1 = 1/a
Rated current
I = S/V
Referred impedance
Z1′ = a²·ZL
Copper loss at part load
Pcu = x²·Pcu,fl
Efficiency
η = x·S·pf/(x·S·pf + Pcore + x²·Pcu,fl)
Maximum efficiency
x = √(Pcore/Pcu,fl)

AC power & power factor

Real power
P = V·I·cos φ
Reactive power
Q = V·I·sin φ
Apparent power
S = V·I = √(P² + Q²)
Power factor
pf = cos φ = P/S
Correction capacitor
QC = P(tan φ1 − tan φ2), C = QC/(ωV²)

AC impedance & phasors

Impedances
ZR = R, ZL = jωL, ZC = 1/(jωC)
Angular frequency
ω = 2πf
Magnitude and angle
|Z| = √(R² + X²), φ = atan(X/R)
Phasor Ohm’s law
V = I·Z

Method

  1. Convert the frequency: ω = 2πf.
  2. Find each reactance: XL = ωL and XC = 1/(ωC).
  3. Add impedances like resistors, with Z = R + jX.
  4. Find the magnitude |Z| = √(R² + X²) and the angle φ = atan(X/R), then use Ohm’s law with phasors.

DC motors

Armature circuit
V = E + IaRa
Back-EMF and torque
E = k·ω, T = k·Ia
Speed
ω = 2π·n/60, ω = (V − IaRa)/k
No-load speed and stall torque
ω0 = V/k, Tstall = k·V/Ra
Powers
Pin = V·Ia, Pdev = E·Ia = T·ω, Pout = Pdev − Prot
Starting
Istart = V/(Ra + Rst)
Speed regulation
SR = (n0 − nFL)/nFL
Maximum developed power
Pmax = V²/(4Ra)
Chopper drive
Va = D·Vdc

Method

  1. Draw the armature circuit: supply V, resistance Ra and the back-EMF E in series.
  2. Use V = E + Ia·Ra to link voltage, current and back-EMF.
  3. Use E = k·ω for speed and T = k·Ia for torque, with ω = 2π·n/60.
  4. For power and efficiency, start from V·Ia, subtract the copper loss Ia²·Ra and then the rotational losses.

2Practice problems

1.Three-phase powerMedium
ABCVL​ = 415 VIL​ = 150 Acos φ = 0.7lagging

A balanced three-phase load draws a line current of 150 A from a supply with a line-to-line voltage of 415 V. The power factor is 0.7 lagging. Find the total real power P drawn by the load in kW.

VL = 415 VIL = 150 Acos φ = 0.7
P = kW
2.Three-phase powerHard
P = 300 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 300 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 0.98 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 300 kWpf1 = 0.75pf2 = 0.98VL = 480 Vf = 50 Hz
Cph = µF
3.TransformersMedium
7200 V240 VIHV​ILV​ = ?500 kVA

A single-phase 500 kVA, 7200/240 V transformer operates at its rated load. Treat it as ideal. Find the current ILV in the low-voltage winding.

S = 500 kVAV1 = 7200 VV2 = 240 V
ILV = A
4.TransformersHard
250 kVA

A single-phase 250 kVA transformer has a core loss of 750 W and a copper loss of 1500 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?

S = 250 kVAPcore = 750 WPcu,fl = 1500 W
Sη,max = kVA
5.AC power & power factorMedium
240 V14 kW, 0.7527 kW, 0.8I

Two loads are connected in parallel to a 240 V single-phase supply: a 4 kW load at a power factor of 0.75 lagging and a 7 kW load at a power factor of 0.8 lagging. Find the overall power factor cos φ seen by the supply.

V = 240 VP1 = 4 kW, pf1 = 0.75 laggingP2 = 7 kW, pf2 = 0.8 lagging
cos φ =
6.AC power & power factorHard
P = 20 kWQ1​S1​φ1​QC​

A single-phase load takes 20 kW at a power factor of 0.7 lagging from a 240 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the new line current Inew.

P = 20 kWpf1 = 0.7pf2 = 0.98V = 240 Vf = 60 Hz
Inew = A
7.AC impedance & phasorsMedium
100 V1000 HzR = 80 ΩL = 200 mHC = 2.2 µF

A series RLC circuit is connected to a 100 V (rms), 1000 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 100 V (rms)f = 1000 HzR = 80 ΩL = 200 mHC = 2.2 µF
φ = °
8.AC impedance & phasorsHard
50 V200 HzR = 80 ΩL = 150 mHC = 1 µF

A series RLC circuit is connected to a 50 V (rms), 200 Hz supply. Find the resonant frequency f0 of the circuit.

V = 50 V (rms)f = 200 HzR = 80 ΩL = 150 mHC = 1 µF
f0 = Hz
9.DC motorsMedium
T (N·m)n (rpm)18 N·mn0​n = ?

A DC motor (k = 0.6 N·m/A, Ra = 0.2 Ω) runs from 120 V and drives a load torque of 18 N·m. Find its speed n in rpm.

V = 120 Vk = 0.6 N·m/ARa = 0.2 ΩT = 18 N·m
n = rpm
10.DC motorsHard
+−220 VD = ?Ra​ = 0.5 ΩM+−En = 900 rpmT = 2.5 N·m

A DC chopper feeds a motor (k = 0.25 V·s/rad, Ra = 0.5 Ω) from a 220 V DC bus. The average armature voltage is D·220 V. Find the duty cycle D that runs the motor at 900 rpm with a load of 2.5 N·m.

Vdc = 220 Vk = 0.25 V·s/radRa = 0.5 ΩT = 2.5 N·mn = 900 rpm
D = %

3Solutions

1.Three-phase power

Answer: P = 75.47 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.

Apparent power

S = √3·VL·IL = √3·415·150 = 107.8 kVA

Real power

P = S·cos φ = 107.8·0.7 = 75.47 kW

Answer

P = 75.47 kW
2.Three-phase power

Answer: Cph = 2814 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.

Reactive power before and after

The real power is unchanged, so each power factor fixes the reactive power: Q = P·tan φ.

Q1 = 300·tan(acos 0.75) = 300·0.8819 = 264.6 kvar
Q2 = 300·tan(acos 0.98) = 300·0.2031 = 60.92 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 203.7 kvar

Capacitance per phase

In a Y bank each capacitor is connected between a line and the star point.

V = 480/√3 = 277.1 V, ω = 2π·50 = 314.2 rad/s
Cph = (QC/3)/(ω·V²) = 67886/(314.2·277.1²) = 2814 µF

Answer

Cph = 2814 µF
3.Transformers

Answer: ILV = 2083 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.

Rated current

Both windings carry the same apparent power.

ILV = S/V = 500000/240 = 2083 A

Answer

ILV = 2083 A
4.Transformers

Answer: Sη,max = 176.8 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.

Copper loss equals core loss

x²·1500 = 750 → x = √(750/1500) = 0.7071

Load at maximum efficiency

Sη,max = x·S = 0.7071·250 = 176.8 kVA

Answer

Sη,max = 176.8 kVA
5.AC power & power factor

Answer: cos φ = 0.7816

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.

Reactive power of each load

A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).

Q1 = 4·tan(arccos 0.75) = 3.528 kvar
Q2 = 7·tan(arccos 0.8) = 5.25 kvar

Totals

P = 4 + 7 = 11 kW
Q = 3.528 + 5.25 = 8.778 kvar
S = √(11² + 8.778²) = 14.07 kVA

Overall power factor

cos φ = P/S = 11/14.07 = 0.7816 lagging

Answer

cos φ = 0.7816
6.AC power & power factor

Answer: Inew = 85.03 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.

Reactive power before and after

The real power P does not change; only the reactive power drawn from the supply does.

Q1 = P·tan φ1 = 20·tan(45.57°) = 20.4 kvar
Q2 = P·tan φ2 = 20·tan(11.48°) = 4.061 kvar
QC = Q1 − Q2 = 16.34 kvar

New line current

Inew = P/(V·cos φ2) = 20000/(240·0.98) = 85.03 A

Answer

Inew = 85.03 A
7.AC impedance & phasors

Answer: φ = 86.14 °

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|.

Reactances

ω = 2πf = 2π·1000 = 6283 rad/s
XL = ωL = 6283·0.2 = 1257 Ω
XC = 1/(ωC) = 1/(6283·2.2·10⁻⁶) = 72.34 Ω

Impedance

In series the impedances add as complex numbers: Z = R + j(XL − XC).

|Z| = √(R² + (XL − XC)²) = √(80² + (1184)²) = 1187 Ω

Phase angle

φ = arctan((XL − XC)/R) = arctan(1184/80) = 86.14°

Answer

φ = 86.14°
8.AC impedance & phasors

Answer: f0 = 410.9 Hz

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|.

Resonance

At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC). The supply frequency does not matter.

f0 = 1/(2π√(LC)) = 1/(2π·√(0.15·1·10⁻⁶)) = 410.9 Hz

Answer

f0 = 410.9 Hz
9.DC motors

Answer: n = 1814 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.

Armature current

Ia = T/k = 18/0.6 = 30 A

Back-EMF and speed

E = 120 − 30·0.2 = 114 V
ω = E/k = 190 rad/s → n = 1814 rpm

Answer

n = 1814 rpm
10.DC motors

Answer: D = 12.98 %

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.

Required armature voltage

Va = 0.25·94.25 + 10·0.5 = 28.56 V

Duty cycle

D = 28.56/220 = 12.98 %

Answer

D = 12.98 %

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