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​ = 10 Acos φ = 0.85lagging

A balanced three-phase load draws a line current of 10 A from a supply with a line-to-line voltage of 208 V. The power factor is 0.85 lagging. Find the total real power P drawn by the load in kW.

VL = 208 VIL = 10 Acos φ = 0.85
P = kW
2.Three-phase powerHard
P = 250 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 250 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 0.98 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 250 kWpf1 = 0.72pf2 = 0.98VL = 690 Vf = 50 Hz
QC = kvar
3.TransformersMedium
6600 V230 VIHV​ILV​ = ?25 kVA

A single-phase 25 kVA, 6600/230 V transformer operates at its rated load. Treat it as ideal. Find the current ILV in the low-voltage winding.

S = 25 kVAV1 = 6600 VV2 = 230 V
ILV = A
4.TransformersHard
pf 0.8575 kVA, 50% load

A single-phase 75 kVA transformer has a core loss of 600 W and a copper loss of 1800 W at rated load. It supplies 50% of its rated kVA at a power factor of 0.85 lagging. Find the total losses Ploss in watts at this load.

S = 75 kVAPcore = 600 WPcu,fl = 1800 Wx = 0.5pf = 0.85
Ploss = W
5.AC power & power factorMedium
P = 1 kWQSφ

A single-phase motor takes 1 kW at a power factor of 0.85 lagging from a 240 V supply. Find the line current I.

P = 1 kWpf = 0.85V = 240 V
I = A
6.AC power & power factorHard
P = 45 kWQ1​S1​φ1​QC​

A single-phase load takes 45 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 = 45 kWpf1 = 0.6pf2 = 0.9V = 240 Vf = 50 Hz
C = µF
7.AC impedance & phasorsMedium
120 V200 HzR = 80 ΩL = 150 mHC = 4.7 µFI

A series RLC circuit is connected to a 120 V (rms), 200 Hz supply. Find the rms current I.

V = 120 V (rms)f = 200 HzR = 80 ΩL = 150 mHC = 4.7 µF
I = A
8.AC impedance & phasorsHard
24 V100 HzR = 15 ΩL = 50 mHC = 10 µFI

A series RLC circuit is connected to a 24 V (rms), 100 Hz supply. Find the rms current I.

V = 24 V (rms)f = 100 HzR = 15 ΩL = 50 mHC = 10 µF
I = A
9.DC motorsMedium
+−VRa​M+−En = 1800 rpmT = ?

A DC motor delivers 11 kW of mechanical power at 1800 rpm. Find the shaft torque T.

P = 11 kWn = 1800 rpm
T = N·m
10.DC motorsHard
+−120 VRs​ = ?Ra​ = 0.4 ΩM+−En = 1500 rpmT = 4 N·m

A DC motor (k = 0.4 V·s/rad, Ra = 0.4 Ω) on a fixed 120 V supply drives a load of 4 N·m. Find the series resistance Rs that slows it down to 1500 rpm.

V = 120 Vk = 0.4 V·s/radRa = 0.4 ΩT = 4 N·mn = 1500 rpm
Rs = Ω

3Solutions

1.Three-phase power

Answer: P = 3.062 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 = √3·208·10 = 3.603 kVA

Real power

P = S·cos φ = 3.603·0.85 = 3.062 kW

Answer

P = 3.062 kW
2.Three-phase power

Answer: QC = 190.2 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 = 250·tan(acos 0.72) = 250·0.9639 = 241 kvar
Q2 = 250·tan(acos 0.98) = 250·0.2031 = 50.76 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 190.2 kvar

Answer

QC = 190.2 kvar
3.Transformers

Answer: ILV = 108.7 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 = 25000/230 = 108.7 A

Answer

ILV = 108.7 A
4.Transformers

Answer: Ploss = 1050 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.

Copper loss at this load

Pcu = x²·Pcu,fl = 0.25·1800 = 450 W

Total losses

The core loss is the same at any load.

Ploss = 600 + 450 = 1050 W

Answer

Ploss = 1050 W

The efficiency at this load is η = 96.81%.

5.AC power & power factor

Answer: I = 4.902 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) = 1000/(240·0.85) = 4.902 A

Answer

I = 4.902 A
6.AC power & power factor

Answer: C = 2111 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 = 45·tan(53.13°) = 60 kvar
Q2 = P·tan φ2 = 45·tan(25.84°) = 21.79 kvar
QC = Q1 − Q2 = 38.21 kvar

Capacitance

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

C = QC/(2πf·V²) = 38206/(2π·50·240²) = 2111 µF

Answer

C = 2111 µF
7.AC impedance & phasors

Answer: I = 1.459 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·200 = 1257 rad/s
XL = ωL = 1257·0.15 = 188.5 Ω
XC = 1/(ωC) = 1/(1257·4.7·10⁻⁶) = 169.3 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(80² + (19.18)²) = 82.27 Ω

Current

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

I = V/|Z| = 120/82.27 = 1.459 A

Answer

I = 1.459 A
8.AC impedance & phasors

Answer: I = 0.1866 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 = 628.3·0.05 = 31.42 Ω
XC = 1/(ωC) = 1/(628.3·10·10⁻⁶) = 159.2 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(15² + (−127.7)²) = 128.6 Ω

Current

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

I = V/|Z| = 24/128.6 = 0.1866 A

Answer

I = 0.1866 A
9.DC motors

Answer: T = 58.36 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·1800/60 = 188.5 rad/s

Torque

T = P/ω = 11000/188.5 = 58.36 N·m

Answer

T = 58.36 N·m
10.DC motors

Answer: Rs = 5.317 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.

Current and back-EMF

Ia = 4/0.4 = 10 A
E = 0.4·157.1 = 62.83 V

Series resistance

Ra + Rs = (120 − 62.83)/10 = 5.717 Ω
Rs = 5.317 Ω

Answer

Rs = 5.317 Ω

The resistor wastes 531.7 W; a chopper controls the speed with far less loss.

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