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​ = 208 VIL​ = 12 Acos φ = 0.7lagging

A balanced three-phase load draws a line current of 12 A from a supply with a line-to-line voltage of 208 V. The power factor is 0.7 lagging. Find the total real power P drawn by the load in kW.

VL = 208 VIL = 12 Acos φ = 0.7
P = kW
2.Three-phase powerHard
P = 80 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 80 kW at a power factor of 0.72 lagging from a 690 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.9 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 80 kWpf1 = 0.72pf2 = 0.9VL = 690 Vf = 50 Hz
QC = kvar
3.TransformersMedium
2400 V240 VIHV​ILV​ = ?500 kVA

A single-phase 500 kVA, 2400/240 V transformer operates at its rated load. Treat it as ideal. Find the current ILV in the low-voltage winding.

S = 500 kVAV1 = 2400 VV2 = 240 V
ILV = A
4.TransformersHard
pf 0.850 kVA, η = 96.5%

A single-phase 50 kVA transformer has a core loss of 200 W. At rated load and a power factor of 0.8 lagging its efficiency is 96.5%. Find the full-load copper loss Pcu,fl in watts.

S = 50 kVAPcore = 200 Wpf = 0.8η = 96.5%
Pcu,fl = W
5.AC power & power factorMedium
P = 10 kWQSφ

A single-phase motor takes 10 kW at a power factor of 0.85 lagging from a 400 V supply. Find the line current I.

P = 10 kWpf = 0.85V = 400 V
I = A
6.AC power & power factorHard
400 V60 Hzload14 kW, 0.8C150 µFI

A single-phase load takes 14 kW at a power factor of 0.8 lagging from a 400 V, 60 Hz supply. A 150 µF capacitor is connected in parallel with the load. Find the new power factor cos φ seen by the supply.

P = 14 kWpf1 = 0.8V = 400 Vf = 60 HzC = 150 µF
cos φ =
7.AC impedance & phasorsMedium
10 V100 HzR = 20 ΩL = 200 mHC = 4.7 µF

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).

V = 10 V (rms)f = 100 HzR = 20 ΩL = 200 mHC = 4.7 µF
φ = °
8.AC impedance & phasorsHard
12 V100 HzR = 80 ΩL = 100 mHC = 22 µFI

A series RLC circuit is connected to a 12 V (rms), 100 Hz supply. Find the rms current I.

V = 12 V (rms)f = 100 HzR = 80 ΩL = 100 mHC = 22 µF
I = A
9.DC motorsMedium
T (N·m)n (rpm)3.2 N·mn0​nFL​

A DC motor (k = 0.8 V·s/rad, Ra = 1 Ω) runs from 120 V. At full load it draws 4 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.

V = 120 Vk = 0.8 V·s/radRa = 1 ΩIFL = 4 A
SR = %
10.DC motorsHard
+−120 VRa​ = 1 ΩIa​: 4 → 8 AM+−En = ?T

A separately excited DC motor on 120 V (Ra = 1 Ω) runs at 1800 rpm with an armature current of 4 A. The load changes and the current becomes 8 A, with the field unchanged. Find the new speed n2.

V = 120 VRa = 1 Ωn1 = 1800 rpmIa1 = 4 AIa2 = 8 A
n2 = rpm

3Solutions

1.Three-phase power

Answer: P = 3.026 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 = √3·208·12 = 4.323 kVA

Real power

P = S·cos φ = 4.323·0.7 = 3.026 kW

Answer

P = 3.026 kW
2.Three-phase power

Answer: QC = 38.36 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 = 80·tan(acos 0.72) = 80·0.9639 = 77.11 kvar
Q2 = 80·tan(acos 0.9) = 80·0.4843 = 38.75 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 38.36 kvar

Answer

QC = 38.36 kvar
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: Pcu,fl = 1251 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.

Output and input power

Pout = 50·0.8 = 40 kW
Pin = 40000/0.965 = 41451 W

Total and copper losses

Ploss = 41451 − 40000 = 1451 W
Pcu,fl = 1451 − 200 = 1251 W

Answer

Pcu,fl = 1251 W
5.AC power & power factor

Answer: I = 29.41 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.

Line current

Only the real part of the power does work, so divide by the power factor.

I = P/(V·cos φ) = 10000/(400·0.85) = 29.41 A

Answer

I = 29.41 A
6.AC power & power factor

Answer: cos φ = 0.9947

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 = 14·tan(arccos 0.8) = 10.5 kvar
QC = V²·2πf·C = 400²·2π·60·150·10⁻⁶ = 9048 var = 9.048 kvar
Q2 = 10.5 − 9.048 = 1.452 kvar

New apparent power

S2 = √(14² + 1.452²) = 14.08 kVA

New power factor

cos φ = P/S2 = 14/14.08 = 0.9947

Answer

cos φ = 0.9947
7.AC impedance & phasors

Answer: φ = −84.63 °

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·100 = 628.3 rad/s
XL = ωL = 628.3·0.2 = 125.7 Ω
XC = 1/(ωC) = 1/(628.3·4.7·10⁻⁶) = 338.6 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(20² + (−213)²) = 213.9 Ω

Phase angle

φ = arctan((XL − XC)/R) = arctan(−213/20) = −84.63°

Answer

φ = −84.63°
8.AC impedance & phasors

Answer: I = 0.149 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|.

Reactances

ω = 2πf = 2π·100 = 628.3 rad/s
XL = ωL = 628.3·0.1 = 62.83 Ω
XC = 1/(ωC) = 1/(628.3·22·10⁻⁶) = 72.34 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(80² + (−9.511)²) = 80.56 Ω

Current

The same current flows through every element of a series circuit.

I = V/|Z| = 12/80.56 = 0.149 A

Answer

I = 0.149 A
9.DC motors

Answer: SR = 3.448 %

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.

No-load and full-load speed

n₀ = 60·120/(2π·0.8) = 1432 rpm
nFL = 60·(120 − 4·1)/(2π·0.8) = 1385 rpm

Speed regulation

SR = (1432 − 1385)/1385·100 = 3.448 %

Answer

SR = 3.448 %
10.DC motors

Answer: n2 = 1738 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.

Back-EMF before and after

E1 = 120 − 4·1 = 116 V
E2 = 120 − 8·1 = 112 V

New speed

n2 = 1800·112/116 = 1738 rpm

Answer

n2 = 1738 rpm

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