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 VIL​ = 75 Acos φ = 0.8lagging

A balanced three-phase load draws a line current of 75 A from a supply with a line-to-line voltage of 400 V. The power factor is 0.8 lagging. Find the total real power P drawn by the load in kW.

VL = 400 VIL = 75 Acos φ = 0.8
P = kW
2.Three-phase powerHard
P = 100 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 100 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. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 100 kWpf1 = 0.7pf2 = 0.92VL = 400 Vf = 50 Hz
QC = kvar
3.TransformersMedium
2400 V240 VIHV​ILV​ = ?25 kVA

A single-phase 25 kVA, 2400/240 V transformer operates at its rated load. Treat it as ideal. Find the current ILV in the low-voltage winding.

S = 25 kVAV1 = 2400 VV2 = 240 V
ILV = A
4.TransformersHard
pf 0.95200 kVA, η = 98.2%

A single-phase 200 kVA transformer has a core loss of 1200 W. At rated load and a power factor of 0.95 lagging its efficiency is 98.2%. Find the full-load copper loss Pcu,fl in watts.

S = 200 kVAPcore = 1200 Wpf = 0.95η = 98.2%
Pcu,fl = W
5.AC power & power factorMedium
P = 36 kWQ = ?Sφ

A single-phase motor takes 36 kW at a power factor of 0.7 lagging from a 240 V supply. Find the reactive power Q it draws.

P = 36 kWpf = 0.7V = 240 V
Q = kvar
6.AC power & power factorHard
120 V50 HzRs​ = 10 ΩXs​ = j30 ΩZL​ = ?

A 120 V (rms), 50 Hz source has an internal impedance Zs = 10 + j30 Ω. The load impedance ZL can be chosen freely. Find the maximum real power Pmax that can be delivered to the load.

Vs = 120 V (rms)f = 50 HzZs = 10 + j30 Ω
Pmax = W
7.AC impedance & phasorsMedium
100 V500 HzR = 10 ΩL = 50 mHC = 4.7 µFI

A series RLC circuit is connected to a 100 V (rms), 500 Hz supply. Find the rms current I.

V = 100 V (rms)f = 500 HzR = 10 ΩL = 50 mHC = 4.7 µF
I = A
8.AC impedance & phasorsHard
10 V100 HzR = 60 ΩL = 50 mHC = 10 µF

A series RLC circuit is connected to a 10 V (rms), 100 Hz supply. Find the resonant frequency f0 of the circuit.

V = 10 V (rms)f = 100 HzR = 60 ΩL = 50 mHC = 10 µF
f0 = Hz
9.DC motorsMedium
+−VRa​M+−En = 3000 rpmT = ?

A DC motor delivers 5.5 kW of mechanical power at 3000 rpm. Find the shaft torque T.

P = 5.5 kWn = 3000 rpm
T = N·m
10.DC motorsHard
+−110 VRa​ = 0.25 ΩM+−En = ?T = 20 N·m

A separately excited DC motor (k = 0.8 V·s/rad at full field, Ra = 0.25 Ω, V = 110 V) drives a constant load torque of 20 N·m. The field flux is reduced to 85 % of its full value. Find the new speed n.

V = 110 Vk = 0.8 V·s/radRa = 0.25 ΩT = 20 N·mΦ = 85 %
n = rpm

3Solutions

1.Three-phase power

Answer: P = 41.57 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 = √3·400·75 = 51.96 kVA

Real power

P = S·cos φ = 51.96·0.8 = 41.57 kW

Answer

P = 41.57 kW
2.Three-phase power

Answer: QC = 59.42 kvar

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 = 100·tan(acos 0.7) = 100·1.02 = 102 kvar
Q2 = 100·tan(acos 0.92) = 100·0.426 = 42.6 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 59.42 kvar

Answer

QC = 59.42 kvar
3.Transformers

Answer: ILV = 104.2 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.

ILV = S/V = 25000/240 = 104.2 A

Answer

ILV = 104.2 A
4.Transformers

Answer: Pcu,fl = 2283 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 = 200·0.95 = 190 kW
Pin = 190000/0.982 = 193483 W

Total and copper losses

Ploss = 193483 − 190000 = 3483 W
Pcu,fl = 3483 − 1200 = 2283 W

Answer

Pcu,fl = 2283 W
5.AC power & power factor

Answer: Q = 36.73 kvar

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.

Angle

φ = arccos(0.7) = 45.57°, tan φ = 1.02

Reactive power

Q = P·tan φ = 36·1.02 = 36.73 kvar

Answer

Q = 36.73 kvar
6.AC power & power factor

Answer: Pmax = 360 W

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.

Conjugate match

The load reactance cancels the source reactance, and the load resistance equals the source resistance.

ZL = Zs* = 10 − j30 Ω

Maximum power

I = Vs/(2Rs) = 120/20 = 6 A
Pmax = I²·RL = Vs²/(4Rs) = 120²/(4·10) = 360 W

Answer

Pmax = 360 W
7.AC impedance & phasors

Answer: I = 1.112 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·500 = 3142 rad/s
XL = ωL = 3142·0.05 = 157.1 Ω
XC = 1/(ωC) = 1/(3142·4.7·10⁻⁶) = 67.73 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(10² + (89.35)²) = 89.91 Ω

Current

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

I = V/|Z| = 100/89.91 = 1.112 A

Answer

I = 1.112 A
8.AC impedance & phasors

Answer: f0 = 225.1 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|.

Resonance

At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC). The supply frequency does not matter.

f0 = 1/(2π√(LC)) = 1/(2π·√(0.05·10·10⁻⁶)) = 225.1 Hz

Answer

f0 = 225.1 Hz
9.DC motors

Answer: T = 17.51 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·3000/60 = 314.2 rad/s

Torque

T = P/ω = 5500/314.2 = 17.51 N·m

Answer

T = 17.51 N·m
10.DC motors

Answer: n = 1441 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 = 0.85·0.8 = 0.68 V·s/rad

New armature current

Ia = 20/0.68 = 29.41 A

New speed

ω = (110 − 29.41·0.25)/0.68 = 151 rad/s → n = 1441 rpm

Answer

n = 1441 rpm

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

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