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​ = 400 VIL​ = 120 Acos φ = 0.9lagging

A balanced three-phase load draws a line current of 120 A from a supply with a line-to-line voltage of 400 V. The power factor is 0.9 lagging. Find the total real power P drawn by the load in kW.

VL = 400 VIL = 120 Acos φ = 0.9
P = kW
2.Three-phase powerHard
P = 50 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 50 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 0.9 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 50 kWpf1 = 0.78pf2 = 0.9VL = 400 Vf = 60 Hz
QC = kvar
3.TransformersMedium
4800 V240 VIHV​ILV​ = ?25 kVA

A single-phase 25 kVA, 4800/240 V transformer operates at its rated load. Treat it as ideal. Find the current ILV in the low-voltage winding.

S = 25 kVAV1 = 4800 VV2 = 240 V
ILV = A
4.TransformersHard
100 kVA

A single-phase 100 kVA transformer has a core loss of 300 W and a copper loss of 450 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?

S = 100 kVAPcore = 300 WPcu,fl = 450 W
Sη,max = kVA
5.AC power & power factorMedium
240 V110 kW, 0.75223 kW, 1I = ?

Two loads are connected in parallel to a 240 V single-phase supply: a 10 kW load at a power factor of 0.75 lagging and a 23 kW heater (unity power factor). Find the total current I drawn from the supply.

V = 240 VP1 = 10 kW, pf1 = 0.75 laggingP2 = 23 kW, pf2 = 1
I = A
6.AC power & power factorHard
P = 15 kWQ1​S1​φ1​QC​

A single-phase load takes 15 kW at a power factor of 0.6 lagging from a 240 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.9. Find the capacitance C needed.

P = 15 kWpf1 = 0.6pf2 = 0.9V = 240 Vf = 50 Hz
C = µF
7.AC impedance & phasorsMedium
Vf0​ = 5000 HzRL = ?C = 2.2 µF

A series RLC circuit with C = 2.2 µF must resonate at 5000 Hz. Find the inductance L needed.

f0 = 5000 HzC = 2.2 µF
L = mH
8.AC impedance & phasorsHard
100 V500 HzR1​ = 15 ΩR2​ = 100 ΩC = 4.7 µF

A resistor R1 is connected in series with a parallel combination of R2 and a capacitor C, across a 100 V (rms), 500 Hz supply. Find the magnitude of the impedance |Z|.

V = 100 V (rms)f = 500 HzR1 = 15 ΩR2 = 100 ΩC = 4.7 µF
|Z| = Ω
9.DC motorsMedium
+−VRa​M+−En = 2900 rpmT = ?

A DC motor delivers 0.75 kW of mechanical power at 2900 rpm. Find the shaft torque T.

P = 0.75 kWn = 2900 rpm
T = N·m
10.DC motorsHard
+−220 VRa​ = 0.8 ΩM+−En = ?T = 15 N·m

A separately excited DC motor (k = 0.6 V·s/rad at full field, Ra = 0.8 Ω, V = 220 V) drives a constant load torque of 15 N·m. The field flux is reduced to 70 % of its full value. Find the new speed n.

V = 220 Vk = 0.6 V·s/radRa = 0.8 ΩT = 15 N·mΦ = 70 %
n = rpm

3Solutions

1.Three-phase power

Answer: P = 74.82 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 = √3·400·120 = 83.14 kVA

Real power

P = S·cos φ = 83.14·0.9 = 74.82 kW

Answer

P = 74.82 kW
2.Three-phase power

Answer: QC = 15.9 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.

Reactive power before and after

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

Q1 = 50·tan(acos 0.78) = 50·0.8023 = 40.11 kvar
Q2 = 50·tan(acos 0.9) = 50·0.4843 = 24.22 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 15.9 kvar

Answer

QC = 15.9 kvar
3.Transformers

Answer: ILV = 104.2 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.

Rated current

Both windings carry the same apparent power.

ILV = S/V = 25000/240 = 104.2 A

Answer

ILV = 104.2 A
4.Transformers

Answer: Sη,max = 81.65 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

x²·450 = 300 → x = √(300/450) = 0.8165

Load at maximum efficiency

Sη,max = x·S = 0.8165·100 = 81.65 kVA

Answer

Sη,max = 81.65 kVA
5.AC power & power factor

Answer: I = 142.3 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 of each load

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

Q1 = 10·tan(arccos 0.75) = 8.819 kvar
Q2 = 0 = 0 kvar

Totals

P = 10 + 23 = 33 kW
Q = 8.819 + 0 = 8.819 kvar
S = √(33² + 8.819²) = 34.16 kVA

Total current

I = S/V = 34158/240 = 142.3 A

Answer

I = 142.3 A
6.AC power & power factor

Answer: C = 703.8 µF

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 = 15·tan(53.13°) = 20 kvar
Q2 = P·tan φ2 = 15·tan(25.84°) = 7.265 kvar
QC = Q1 − Q2 = 12.74 kvar

Capacitance

A capacitor across the supply delivers Q = V²·ωC.

C = QC/(2πf·V²) = 12735/(2π·50·240²) = 703.8 µF

Answer

C = 703.8 µF
7.AC impedance & phasors

Answer: L = 0.4606 mH

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 condition

At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC).

ω₀ = 2π·5000 = 31416 rad/s
L = 1/(ω₀²·C) = 1/(31416²·2.2·10⁻⁶) = 0.4606 mH

Answer

L = 0.4606 mH
8.AC impedance & phasors

Answer: |Z| = 65.67 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|.

Reactance

XC = 1/(2π·500·4.7·10⁻⁶) = 67.73 Ω

Parallel combination

Multiply top and bottom by the complex conjugate to split Zp into real and imaginary parts.

Zp = R₂·(−jXC)/(R₂ − jXC) = (R₂·XC² − jR₂²·XC)/(R₂² + XC²)
Zp = 31.44 − j46.43 Ω

Total impedance

Z = 15 + 31.44 − j46.43 = 46.44 − j46.43 Ω
|Z| = √(46.44² + 46.43²) = 65.67 Ω

Answer

|Z| = 65.67 Ω
9.DC motors

Answer: T = 2.47 N·m

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.

Speed in rad/s

ω = 2π·2900/60 = 303.7 rad/s

Torque

T = P/ω = 750/303.7 = 2.47 N·m

Answer

T = 2.47 N·m
10.DC motors

Answer: n = 4352 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.

Weakened motor constant

k₂ = 0.7·0.6 = 0.42 V·s/rad

New armature current

Ia = 15/0.42 = 35.71 A

New speed

ω = (220 − 35.71·0.8)/0.42 = 455.8 rad/s → n = 4352 rpm

Answer

n = 4352 rpm

Field weakening is how a DC drive runs above its base speed.

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