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​ = 400 VZ = 18 + j24 Ωper phase

A balanced Δ-connected load with an impedance of 18 + j24 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 400 V. Find the total real power P absorbed by the load in kW.

VL = 400 VZ = 18 + j24 Ω per phase
P = kW
2.Three-phase powerHard
P = 60 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 60 kW at a power factor of 0.78 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 = 60 kWpf1 = 0.78pf2 = 0.92VL = 415 Vf = 60 Hz
Cph = µF
3.TransformersMedium
2400 V240 VIHV​ILV​ = ?50 kVA

A single-phase 50 kVA, 2400/240 V transformer operates at its rated load. Treat it as ideal. Find the current ILV in the low-voltage winding.

S = 50 kVAV1 = 2400 VV2 = 240 V
ILV = A
4.TransformersHard
pf 0.8250 kVA, η = 99%

A single-phase 250 kVA transformer has a core loss of 1000 W. At rated load and a power factor of 0.8 lagging its efficiency is 99%. Find the full-load copper loss Pcu,fl in watts.

S = 250 kVAPcore = 1000 Wpf = 0.8η = 99%
Pcu,fl = W
5.AC power & power factorMedium
120 VZ7 + j24 ΩI

A load with impedance Z = 7 + j24 Ω is connected to a 120 V (rms) single-phase supply. Find the reactive power Q it absorbs.

V = 120 VZ = 7 + j24 Ω
Q = var
6.AC power & power factorHard
P = 25 kWQ1​S1​φ1​QC​

A single-phase load takes 25 kW at a power factor of 0.65 lagging from a 230 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the new line current Inew.

P = 25 kWpf1 = 0.65pf2 = 0.98V = 230 Vf = 60 Hz
Inew = A
7.AC impedance & phasorsMedium
Vf0​ = 1000 HzRL = 100 mHC = ?

A series RLC circuit with L = 100 mH must resonate at 1000 Hz. Find the capacitance C needed.

f0 = 1000 HzL = 100 mH
C = µF
8.AC impedance & phasorsHard
24 V1000 HzR = 10 ΩL = 20 mHC = 100 µF+−VC​

A series RLC circuit is connected to a 24 V (rms), 1000 Hz supply. Find the rms voltage VC across the capacitor.

V = 24 V (rms)f = 1000 HzR = 10 ΩL = 20 mHC = 100 µF
VC = V
9.DC motorsMedium
+−VRa​M+−En = 1450 rpmT = ?

A DC motor delivers 0.5 kW of mechanical power at 1450 rpm. Find the shaft torque T.

P = 0.5 kWn = 1450 rpm
T = N·m
10.DC motorsHard
+−110 VRs​ = ?Ra​ = 0.25 ΩM+−En = 600 rpmT = 12 N·m

A DC motor (k = 0.8 V·s/rad, Ra = 0.25 Ω) on a fixed 110 V supply drives a load of 12 N·m. Find the series resistance Rs that slows it down to 600 rpm.

V = 110 Vk = 0.8 V·s/radRa = 0.25 ΩT = 12 N·mn = 600 rpm
Rs = Ω

3Solutions

1.Three-phase power

Answer: P = 9.6 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 = 400 V

Phase current

|Z| = √(18² + 24²) = 30 Ω
Iph = 400/30 = 13.33 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·13.33²·18 = 9.6 kW

Answer

P = 9.6 kW

Check: cos φ = R/|Z| = 0.6, and √3·VL·IL·cos φ gives the same power.

2.Three-phase power

Answer: Cph = 115.9 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 = 60·tan(acos 0.78) = 60·0.8023 = 48.14 kvar
Q2 = 60·tan(acos 0.92) = 60·0.426 = 25.56 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 22.58 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) = 7526/(377·415²) = 115.9 µF

Answer

Cph = 115.9 µF
3.Transformers

Answer: ILV = 208.3 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 = 50000/240 = 208.3 A

Answer

ILV = 208.3 A
4.Transformers

Answer: Pcu,fl = 1020 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 = 250·0.8 = 200 kW
Pin = 200000/0.99 = 202020 W

Total and copper losses

Ploss = 202020 − 200000 = 2020 W
Pcu,fl = 2020 − 1000 = 1020 W

Answer

Pcu,fl = 1020 W
5.AC power & power factor

Answer: Q = 553 var

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| = √(7² + 24²) = 25 Ω
I = V/|Z| = 120/25 = 4.8 A

Reactive power

Q = I²·X = 4.8²·24 = 553 var

Answer

Q = 553 var
6.AC power & power factor

Answer: Inew = 110.9 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 = 25·tan(49.46°) = 29.23 kvar
Q2 = P·tan φ2 = 25·tan(11.48°) = 5.076 kvar
QC = Q1 − Q2 = 24.15 kvar

New line current

Inew = P/(V·cos φ2) = 25000/(230·0.98) = 110.9 A

Answer

Inew = 110.9 A
7.AC impedance & phasors

Answer: C = 0.2533 µF

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 condition

At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC).

ω₀ = 2π·1000 = 6283 rad/s
C = 1/(ω₀²·L) = 1/(6283²·0.1) = 0.2533 µF

Answer

C = 0.2533 µF
8.AC impedance & phasors

Answer: VC = 0.3069 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·1000 = 6283 rad/s
XL = ωL = 6283·0.02 = 125.7 Ω
XC = 1/(ωC) = 1/(6283·100·10⁻⁶) = 1.592 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(10² + (124.1)²) = 124.5 Ω

Current

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

I = V/|Z| = 24/124.5 = 0.1928 A

Voltage across the capacitor

VC = I·XC = 0.1928·1.592 = 0.3069 V

Answer

VC = 0.3069 V
9.DC motors

Answer: T = 3.293 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 = 500/151.8 = 3.293 N·m

Answer

T = 3.293 N·m
10.DC motors

Answer: Rs = 3.732 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 = 12/0.8 = 15 A
E = 0.8·62.83 = 50.27 V

Series resistance

Ra + Rs = (110 − 50.27)/15 = 3.982 Ω
Rs = 3.732 Ω

Answer

Rs = 3.732 Ω

The resistor wastes 839.8 W; a chopper controls the speed with far less loss.

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