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​ = 480 VIL​ = ?M11 kW, η = 90%cos φ = 0.85

A three-phase induction motor delivers 11 kW of mechanical output power at full load. At this load its efficiency is 90% and its power factor is 0.85 lagging. The supply has a line-to-line voltage of 480 V. Find the line current IL.

Pout = 11 kWη = 90%cos φ = 0.85VL = 480 V
IL = A
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 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 0.95 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 80 kWpf1 = 0.72pf2 = 0.95VL = 400 Vf = 50 Hz
Cph = µF
3.TransformersMedium
75 Ω101

A 75 Ω resistor is connected across the secondary winding of an ideal transformer with a turns ratio N₁:N₂ = 10:1. Find the resistance R1′ seen looking into the primary winding.

N1:N2 = 10:1RL = 75 Ω
R1′ = Ω
4.TransformersHard
25 kVA

A single-phase 25 kVA transformer has a core loss of 150 W and a copper loss of 600 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?

S = 25 kVAPcore = 150 WPcu,fl = 600 W
Sη,max = kVA
5.AC power & power factorMedium
120 VZ15 + j20 ΩI

A load with impedance Z = 15 + j20 Ω is connected to a 120 V (rms) single-phase supply. Find the apparent power S.

V = 120 VZ = 15 + j20 Ω
S = VA
6.AC power & power factorHard
P = 20 kWQ1​S1​φ1​QC​

A single-phase load takes 20 kW at a power factor of 0.75 lagging from a 400 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.9. Find the new line current Inew.

P = 20 kWpf1 = 0.75pf2 = 0.9V = 400 Vf = 50 Hz
Inew = A
7.AC impedance & phasorsMedium
10 V200 HzR = 15 ΩL = 150 mHC = 100 µF

A series RLC circuit is connected to a 10 V (rms), 200 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 10 V (rms)f = 200 HzR = 15 ΩL = 150 mHC = 100 µF
φ = °
8.AC impedance & phasorsHard
VfR = 50 ΩL = 2 mHC = 0.01 µF

A series RLC circuit has R = 50 Ω, L = 2 mH and C = 0.01 µF. Find its bandwidth B (the width of the band between the half-power frequencies), in Hz.

R = 50 ΩL = 2 mHC = 0.01 µF
B = Hz
9.DC motorsMedium
+−VRa​M+−En = 1450 rpmT = ?

A DC motor delivers 11 kW of mechanical power at 1450 rpm. Find the shaft torque T.

P = 11 kWn = 1450 rpm
T = N·m
10.DC motorsHard
+−240 VRs​ = ?Ra​ = 0.8 ΩM+−En = 1000 rpmT = 7.2 N·m

A DC motor (k = 0.6 V·s/rad, Ra = 0.8 Ω) on a fixed 240 V supply drives a load of 7.2 N·m. Find the series resistance Rs that slows it down to 1000 rpm.

V = 240 Vk = 0.6 V·s/radRa = 0.8 ΩT = 7.2 N·mn = 1000 rpm
Rs = Ω

3Solutions

1.Three-phase power

Answer: IL = 17.3 A

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.

Electrical input power

Pin = Pout/η = 11/0.9 = 12.22 kW

Line current

IL = Pin/(√3·VL·cos φ) = 12222/(√3·480·0.85) = 17.3 A

Answer

IL = 17.3 A
2.Three-phase power

Answer: Cph = 337 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 = 80·tan(acos 0.72) = 80·0.9639 = 77.11 kvar
Q2 = 80·tan(acos 0.95) = 80·0.3287 = 26.29 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 50.81 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) = 16938/(314.2·400²) = 337 µF

Answer

Cph = 337 µF
3.Transformers

Answer: R1′ = 7500 Ω

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.

Referred resistance

The voltage goes up by a and the current down by a, so the ratio V/I goes up by a².

R1′ = a²·RL = 10²·75 = 7500 Ω

Answer

R1′ = 7500 Ω
4.Transformers

Answer: Sη,max = 12.5 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

x²·600 = 150 → x = √(150/600) = 0.5

Load at maximum efficiency

Sη,max = x·S = 0.5·25 = 12.5 kVA

Answer

Sη,max = 12.5 kVA
5.AC power & power factor

Answer: S = 576 VA

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.

Current

|Z| = √(15² + 20²) = 25 Ω
I = V/|Z| = 120/25 = 4.8 A

Apparent power

S = V·I = 120·4.8 = 576 VA

Answer

S = 576 VA
6.AC power & power factor

Answer: Inew = 55.56 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(41.41°) = 17.64 kvar
Q2 = P·tan φ2 = 20·tan(25.84°) = 9.686 kvar
QC = Q1 − Q2 = 7.952 kvar

New line current

Inew = P/(V·cos φ2) = 20000/(400·0.9) = 55.56 A

Answer

Inew = 55.56 A
7.AC impedance & phasors

Answer: φ = 85.25 °

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/(1257·100·10⁻⁶) = 7.958 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(15² + (180.5)²) = 181.2 Ω

Phase angle

φ = arctan((XL − XC)/R) = arctan(180.5/15) = 85.25°

Answer

φ = 85.25°
8.AC impedance & phasors

Answer: B = 3979 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|.

Resonant frequency

f0 = 1/(2π·√(0.002·0.01·10⁻⁶)) = 35588 Hz

Quality factor

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

Q = (1/R)·√(L/C) = (1/50)·√(0.002/(0.01·10⁻⁶)) = 8.944

Bandwidth

B = f0/Q = 35588/8.944 = 3979 Hz

Answer

B = 3979 Hz
9.DC motors

Answer: T = 72.44 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·1450/60 = 151.8 rad/s

Torque

T = P/ω = 11000/151.8 = 72.44 N·m

Answer

T = 72.44 N·m
10.DC motors

Answer: Rs = 13.96 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 = 7.2/0.6 = 12 A
E = 0.6·104.7 = 62.83 V

Series resistance

Ra + Rs = (240 − 62.83)/12 = 14.76 Ω
Rs = 13.96 Ω

Answer

Rs = 13.96 Ω

The resistor wastes 2011 W; a chopper controls the speed with far less loss.

ExamProblems · examproblems.com/courses/power-systems