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​ = 690 VIL​ = ?M55 kW, η = 91%cos φ = 0.85

A three-phase induction motor delivers 55 kW of mechanical output power at full load. At this load its efficiency is 91% and its power factor is 0.85 lagging. The supply has a line-to-line voltage of 690 V. Find the line current IL.

Pout = 55 kWη = 91%cos φ = 0.85VL = 690 V
IL = A
2.Three-phase powerHard
P = 75 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 75 kW at a power factor of 0.75 lagging from a 690 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.92 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 = 0.75pf2 = 0.92VL = 690 Vf = 60 Hz
Cph = µF
3.TransformersMedium
8 ΩN1​N2​1152 Ω

An amplifier with an output resistance of 1152 Ω drives an 8 Ω loudspeaker through an ideal transformer. For maximum power transfer the loudspeaker must look like 1152 Ω from the amplifier. Find the turns ratio a = N₁/N₂.

Rs = 1152 ΩRL = 8 Ω
a =
4.TransformersHard
pf 0.8200 kVA, η = 96.5%

A single-phase 200 kVA transformer has a core loss of 800 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 = 200 kVAPcore = 800 Wpf = 0.8η = 96.5%
Pcu,fl = W
5.AC power & power factorMedium
240 V16 kW, 0.7214 kW, 1I = ?

Two loads are connected in parallel to a 240 V single-phase supply: a 6 kW load at a power factor of 0.7 lagging and a 14 kW heater (unity power factor). Find the total current I drawn from the supply.

V = 240 VP1 = 6 kW, pf1 = 0.7 laggingP2 = 14 kW, pf2 = 1
I = A
6.AC power & power factorHard
P = 30 kWQ1​S1​φ1​QC​

A single-phase load takes 30 kW at a power factor of 0.7 lagging from a 400 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.92. Find the capacitance C needed.

P = 30 kWpf1 = 0.7pf2 = 0.92V = 400 Vf = 60 Hz
C = µF
7.AC impedance & phasorsMedium
100 V500 HzR = 30 ΩL = 50 mHC = 10 µF

A series RLC circuit is connected to a 100 V (rms), 500 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 100 V (rms)f = 500 HzR = 30 ΩL = 50 mHC = 10 µF
φ = °
8.AC impedance & phasorsHard
100 V100 HzR = 20 ΩL = 250 mHC = 47 µF+−VC​

A series RLC circuit is connected to a 100 V (rms), 100 Hz supply. Find the rms voltage VC across the capacitor.

V = 100 V (rms)f = 100 HzR = 20 ΩL = 250 mHC = 47 µF
VC = V
9.DC motorsMedium
T (N·m)n (rpm)0.5 N·mn0​nFL​

A DC motor (k = 0.25 V·s/rad, Ra = 2 Ω) runs from 48 V. At full load it draws 2 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.

V = 48 Vk = 0.25 V·s/radRa = 2 ΩIFL = 2 A
SR = %
10.DC motorsHard
T (N·m)n (rpm)50 N·m500 rpm

A DC motor with k = 2 V·s/rad and Ra = 1 Ω must drive a load of 50 N·m at 500 rpm. Find the armature voltage V it needs.

k = 2 V·s/radRa = 1 ΩT = 50 N·mn = 500 rpm
V = V

3Solutions

1.Three-phase power

Answer: IL = 59.5 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 = 55/0.91 = 60.44 kW

Line current

IL = Pin/(√3·VL·cos φ) = 60440/(√3·690·0.85) = 59.5 A

Answer

IL = 59.5 A
2.Three-phase power

Answer: Cph = 63.5 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 0.75) = 75·0.8819 = 66.14 kvar
Q2 = 75·tan(acos 0.92) = 75·0.426 = 31.95 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 34.19 kvar

Capacitance per phase

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

V = 690 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 11398/(377·690²) = 63.5 µF

Answer

Cph = 63.5 µF
3.Transformers

Answer: a = 12

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.

Match the referred load

a²·8 = 1152
a = √(1152/8) = √144 = 12

Answer

a = 12
4.Transformers

Answer: Pcu,fl = 5003 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 = 200·0.8 = 160 kW
Pin = 160000/0.965 = 165803 W

Total and copper losses

Ploss = 165803 − 160000 = 5803 W
Pcu,fl = 5803 − 800 = 5003 W

Answer

Pcu,fl = 5003 W
5.AC power & power factor

Answer: I = 87.15 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 = 6·tan(arccos 0.7) = 6.121 kvar
Q2 = 0 = 0 kvar

Totals

P = 6 + 14 = 20 kW
Q = 6.121 + 0 = 6.121 kvar
S = √(20² + 6.121²) = 20.92 kVA

Total current

I = S/V = 20916/240 = 87.15 A

Answer

I = 87.15 A
6.AC power & power factor

Answer: C = 295.5 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 = 30·tan(45.57°) = 30.61 kvar
Q2 = P·tan φ2 = 30·tan(23.07°) = 12.78 kvar
QC = Q1 − Q2 = 17.83 kvar

Capacitance

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

C = QC/(2πf·V²) = 17826/(2π·60·400²) = 295.5 µF

Answer

C = 295.5 µF
7.AC impedance & phasors

Answer: φ = 76.53 °

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·500 = 3142 rad/s
XL = ωL = 3142·0.05 = 157.1 Ω
XC = 1/(ωC) = 1/(3142·10·10⁻⁶) = 31.83 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(30² + (125.2)²) = 128.8 Ω

Phase angle

φ = arctan((XL − XC)/R) = arctan(125.2/30) = 76.53°

Answer

φ = 76.53°
8.AC impedance & phasors

Answer: VC = 27.13 V

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 = 628.3·0.25 = 157.1 Ω
XC = 1/(ωC) = 1/(628.3·47·10⁻⁶) = 33.86 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(20² + (123.2)²) = 124.8 Ω

Current

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

I = V/|Z| = 100/124.8 = 0.8011 A

Voltage across the capacitor

VC = I·XC = 0.8011·33.86 = 27.13 V

Answer

VC = 27.13 V
9.DC motors

Answer: SR = 9.091 %

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 = 60·48/(2π·0.25) = 1833 rpm
nFL = 60·(48 − 2·2)/(2π·0.25) = 1681 rpm

Speed regulation

SR = (1833 − 1681)/1681·100 = 9.091 %

Answer

SR = 9.091 %
10.DC motors

Answer: V = 129.7 V

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

E = 2·52.36 = 104.7 V

Armature current

Ia = 50/2 = 25 A

Armature voltage

V = 104.7 + 25·1 = 129.7 V

Answer

V = 129.7 V

ExamProblems · examproblems.com/courses/power-systems