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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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​ = 30 Acos φ = 0.85lagging

A balanced three-phase load draws a line current of 30 A from a supply with a line-to-line voltage of 415 V. The power factor is 0.85 lagging. Find the total real power P drawn by the load in kW.

VL = 415 VIL = 30 Acos φ = 0.85
P = kW
2.Three-phase powerHard
P = 75 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 75 kW at a power factor of 0.7 lagging from a 400 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.92 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 75 kWpf1 = 0.7pf2 = 0.92VL = 400 Vf = 50 Hz
Cph = µF
3.TransformersMedium
4160 V480 VIHV​ = ?ILV​500 kVA

A single-phase 500 kVA, 4160/480 V transformer operates at its rated load. Treat it as ideal. Find the current IHV in the high-voltage winding.

S = 500 kVAV1 = 4160 VV2 = 480 V
IHV = A
4.TransformersHard
pf 0.9250 kVA, η = 97%

A single-phase 250 kVA transformer has a core loss of 1000 W. At rated load and a power factor of 0.9 lagging its efficiency is 97%. Find the full-load copper loss Pcu,fl in watts.

S = 250 kVAPcore = 1000 Wpf = 0.9η = 97%
Pcu,fl = W
5.AC power & power factorMedium
240 VR + jXP = 3162 W15.5 A

An inductive load takes 15.5 A and 3162 W from a 240 V (rms) single-phase supply. Model it as a resistance R in series with a reactance X. Find X.

V = 240 VI = 15.5 AP = 3162 W
X = Ω
6.AC power & power factorHard
400 V50 Hzload14 kW, 0.65C220 µFI = ?

A single-phase load takes 14 kW at a power factor of 0.65 lagging from a 400 V, 50 Hz supply. A 220 µF capacitor is connected in parallel with the load. Find the new line current I.

P = 14 kWpf1 = 0.65V = 400 Vf = 50 HzC = 220 µF
I = A
7.AC impedance & phasorsMedium
12 V50 HzR = 20 ΩL = 100 mHC = 22 µFI

A series RLC circuit is connected to a 12 V (rms), 50 Hz supply. Find the rms current I.

V = 12 V (rms)f = 50 HzR = 20 ΩL = 100 mHC = 22 µF
I = A
8.AC impedance & phasorsHard
VfR = 50 ΩL = 200 mHC = 0.022 µF

A series RLC circuit has R = 50 Ω, L = 200 mH and C = 0.022 µF. Find its quality factor Q at resonance.

R = 50 ΩL = 200 mHC = 0.022 µF
Q =
9.DC motorsMedium
T (N·m)n (rpm)30 N·mn0​nFL​

A DC motor (k = 1.2 V·s/rad, Ra = 1 Ω) runs from 220 V. At full load it draws 25 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.

V = 220 Vk = 1.2 V·s/radRa = 1 ΩIFL = 25 A
SR = %
10.DC motorsHard
+−120 VRst​ = ?Ra​ = 0.5 Ω100 AM+−E = 0n = 0T

A 120 V DC motor has an armature resistance of 0.5 Ω. Find the series starting resistance Rst that limits the starting current to 100 A.

V = 120 VRa = 0.5 ΩImax = 100 A
Rst = Ω

3Solutions

1.Three-phase power

Answer: P = 18.33 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 = √3·415·30 = 21.56 kVA

Real power

P = S·cos φ = 21.56·0.85 = 18.33 kW

Answer

P = 18.33 kW
2.Three-phase power

Answer: Cph = 295.5 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 = 75·tan(acos 0.7) = 75·1.02 = 76.52 kvar
Q2 = 75·tan(acos 0.92) = 75·0.426 = 31.95 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 44.57 kvar

Capacitance per phase

In a Δ bank each capacitor is connected between two lines.

V = 400 V, ω = 2π·50 = 314.2 rad/s
Cph = (QC/3)/(ω·V²) = 14855/(314.2·400²) = 295.5 µF

Answer

Cph = 295.5 µF
3.Transformers

Answer: IHV = 120.2 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.

IHV = S/V = 500000/4160 = 120.2 A

Answer

IHV = 120.2 A
4.Transformers

Answer: Pcu,fl = 5959 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 = 250·0.9 = 225 kW
Pin = 225000/0.97 = 231959 W

Total and copper losses

Ploss = 231959 − 225000 = 6959 W
Pcu,fl = 6959 − 1000 = 5959 W

Answer

Pcu,fl = 5959 W
5.AC power & power factor

Answer: X = 8.157 Ω

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.

Resistance from the real power

R = P/I² = 3162/15.5² = 13.16 Ω

Reactance from the impedance

|Z| = V/I = 240/15.5 = 15.48 Ω
X = √(|Z|² − R²) = √(15.48² − 13.16²) = 8.157 Ω

Answer

X = 8.157 Ω
6.AC power & power factor

Answer: I = 37.43 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 = 14·tan(arccos 0.65) = 16.37 kvar
QC = V²·2πf·C = 400²·2π·50·220·10⁻⁶ = 11058 var = 11.06 kvar
Q2 = 16.37 − 11.06 = 5.309 kvar

New apparent power

S2 = √(14² + 5.309²) = 14.97 kVA

New line current

I = S2/V = 14973/400 = 37.43 A

Answer

I = 37.43 A
7.AC impedance & phasors

Answer: I = 0.1043 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·50 = 314.2 rad/s
XL = ωL = 314.2·0.1 = 31.42 Ω
XC = 1/(ωC) = 1/(314.2·22·10⁻⁶) = 144.7 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(20² + (−113.3)²) = 115 Ω

Current

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

I = V/|Z| = 12/115 = 0.1043 A

Answer

I = 0.1043 A
8.AC impedance & phasors

Answer: Q = 60.3

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|.

Resonant frequency

f0 = 1/(2π·√(0.2·0.022·10⁻⁶)) = 2399 Hz

Quality factor

Q compares the reactance at resonance with the resistance: Q = ω₀L/R.

Q = (1/R)·√(L/C) = (1/50)·√(0.2/(0.022·10⁻⁶)) = 60.3

Answer

Q = 60.3
9.DC motors

Answer: SR = 12.82 %

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 = 60·220/(2π·1.2) = 1751 rpm
nFL = 60·(220 − 25·1)/(2π·1.2) = 1552 rpm

Speed regulation

SR = (1751 − 1552)/1552·100 = 12.82 %

Answer

SR = 12.82 %
10.DC motors

Answer: Rst = 0.7 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.

Total resistance

Ra + Rst = 120/100 = 1.2 Ω

Starting resistor

Rst = 1.2 − 0.5 = 0.7 Ω

Answer

Rst = 0.7 Ω

The resistor is cut out step by step as the motor speeds up and the back-EMF grows.

ExamProblems · examproblems.com/courses/power-systems