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 VZ = 12 + j5 Ωper phase

A balanced Δ-connected load with an impedance of 12 + j5 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 415 V. Find the total real power P absorbed by the load in kW.

VL = 415 VZ = 12 + j5 Ω per phase
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 0.8 lagging from a 415 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 = 250 kWpf1 = 0.8pf2 = 0.92VL = 415 Vf = 60 Hz
Cph = µF
3.TransformersMedium
RL​ = 30 Ω41960 VV2​I1​ = ?I2​

An ideal transformer with a turns ratio N₁:N₂ = 4:1 has 960 V applied to its primary winding. A 30 Ω resistor is connected across the secondary winding. Find the primary current I1.

N1:N2 = 4:1V1 = 960 VRL = 30 Ω
I1 = A
4.TransformersHard
pf 0.9525 kVA, η = 96%

A single-phase 25 kVA transformer has a core loss of 150 W. At rated load and a power factor of 0.95 lagging its efficiency is 96%. Find the full-load copper loss Pcu,fl in watts.

S = 25 kVAPcore = 150 Wpf = 0.95η = 96%
Pcu,fl = W
5.AC power & power factorMedium
120 VZ12 + j5 ΩI

A load with impedance Z = 12 + j5 Ω is connected to a 120 V (rms) single-phase supply. Find the apparent power S.

V = 120 VZ = 12 + j5 Ω
S = VA
6.AC power & power factorHard
P1​ = 32 kWQ1​S1​φ

A 75 kVA single-phase transformer supplies a load of 32 kW at a power factor of 0.65 lagging. How much extra real power ΔP can be added as a resistive (unity power factor) load before the transformer reaches its kVA rating?

Smax = 75 kVAP1 = 32 kWpf1 = 0.65
ΔP = kW
7.AC impedance & phasorsMedium
10 V60 HzR = 10 ΩL = 200 mHC = 22 µFI

A series RLC circuit is connected to a 10 V (rms), 60 Hz supply. Find the rms current I.

V = 10 V (rms)f = 60 HzR = 10 ΩL = 200 mHC = 22 µF
I = A
8.AC impedance & phasorsHard
1 Vf0​R = 20 ΩL = 5 mHC = 0.22 µF

A series RLC circuit is driven by a 1 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.

V = 1 V (rms)R = 20 ΩL = 5 mHC = 0.22 µF
VL = V
9.DC motorsMedium
T (N·m)n (rpm)Tst​ = ?n0​

A DC motor with k = 0.75 N·m/A and Ra = 1.2 Ω is connected to 48 V. Find its stall torque Tstall, the torque at zero speed.

V = 48 Vk = 0.75 N·m/ARa = 1.2 Ω
Tstall = N·m
10.DC motorsHard
+−240 VRa​ = 0.2 ΩM+−En = ?T = 80 N·m

A separately excited DC motor (k = 2 V·s/rad at full field, Ra = 0.2 Ω, V = 240 V) drives a constant load torque of 80 N·m. The field flux is reduced to 70 % of its full value. Find the new speed n.

V = 240 Vk = 2 V·s/radRa = 0.2 ΩT = 80 N·mΦ = 70 %
n = rpm

3Solutions

1.Three-phase power

Answer: P = 36.69 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.

Phase voltage

Vph = VL = 415 V

Phase current

|Z| = √(12² + 5²) = 13 Ω
Iph = 415/13 = 31.92 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·31.92²·12 = 36.69 kW

Answer

P = 36.69 kW

Check: cos φ = R/|Z| = 0.9231, and √3·VL·IL·cos φ gives the same power.

2.Three-phase power

Answer: Cph = 415.9 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 = 250·tan(acos 0.8) = 250·0.75 = 187.5 kvar
Q2 = 250·tan(acos 0.92) = 250·0.426 = 106.5 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 81 kvar

Capacitance per phase

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

V = 415 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 27000/(377·415²) = 415.9 µF

Answer

Cph = 415.9 µF
3.Transformers

Answer: I1 = 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.

Secondary voltage and current

V2 = 960/4 = 240 V
I2 = 240/30 = 8 A

Primary current

An ideal transformer has no losses: V₁·I₁ = V₂·I₂.

I1 = I2/4 = 2 A

Answer

I1 = 2 A

Check: the load seen from the primary is a²·RL = 480 Ω, and 960/480 gives the same current.

4.Transformers

Answer: Pcu,fl = 839.6 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 = 25·0.95 = 23.75 kW
Pin = 23750/0.96 = 24740 W

Total and copper losses

Ploss = 24740 − 23750 = 989.6 W
Pcu,fl = 989.6 − 150 = 839.6 W

Answer

Pcu,fl = 839.6 W
5.AC power & power factor

Answer: S = 1108 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| = √(12² + 5²) = 13 Ω
I = V/|Z| = 120/13 = 9.231 A

Apparent power

S = V·I = 120·9.231 = 1108 VA

Answer

S = 1108 VA
6.AC power & power factor

Answer: ΔP = 33 kW

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 the existing load

Q1 = 32·tan(arccos 0.65) = 37.41 kvar

Largest total real power

At full rating S = Smax, and Q is unchanged by the resistive load.

P = √(Smax² − Q1²) = √(75² − 37.41²) = 65 kW

Extra real power

ΔP = 65 − 32 = 33 kW

Answer

ΔP = 33 kW
7.AC impedance & phasors

Answer: I = 0.2161 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·60 = 377 rad/s
XL = ωL = 377·0.2 = 75.4 Ω
XC = 1/(ωC) = 1/(377·22·10⁻⁶) = 120.6 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(10² + (−45.17)²) = 46.27 Ω

Current

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

I = V/|Z| = 10/46.27 = 0.2161 A

Answer

I = 0.2161 A
8.AC impedance & phasors

Answer: VL = 7.538 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|.

Resonance

At resonance the reactances cancel, so the impedance is just R.

ω₀ = 1/√(LC) = 1/√(0.005·0.22·10⁻⁶) = 30151 rad/s
I = V/R = 1/20 = 0.05 A

Voltage across the inductor

XL = ω₀L = 30151·0.005 = 150.8 Ω
VL = I·XL = 0.05·150.8 = 7.538 V

Answer

VL = 7.538 V

This is Q = 7.538 times the supply voltage.

9.DC motors

Answer: Tstall = 30 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.

Stall current

Ia = V/Ra = 48/1.2 = 40 A

Stall torque

Tstall = 0.75·40 = 30 N·m

Answer

Tstall = 30 N·m
10.DC motors

Answer: n = 1559 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.

Weakened motor constant

k₂ = 0.7·2 = 1.4 V·s/rad

New armature current

Ia = 80/1.4 = 57.14 A

New speed

ω = (240 − 57.14·0.2)/1.4 = 163.3 rad/s → n = 1559 rpm

Answer

n = 1559 rpm

Field weakening is how a DC drive runs above its base speed.

ExamProblems · examproblems.com/courses/power-systems