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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
Convert the frequency: ω = 2πf.
Find each reactance: XL = ωL and XC = 1/(ωC).
Add impedances like resistors, with Z = R + jX.
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
Draw the armature circuit: supply V, resistance Ra and the back-EMF E in series.
Use V = E + Ia·Ra to link voltage, current and back-EMF.
Use E = k·ω for speed and T = k·Ia for torque, with ω = 2π·n/60.
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
A balanced Y-connected load with an impedance of 12 + j9 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 480 V. Find the total real power P absorbed by the load in kW.
VL = 480 VZ = 12 + j9 Ω per phase
P = kW
2.Three-phase powerHard
A balanced three-phase load takes 120 kW at a power factor of 0.75 lagging from a 480 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.9 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.
P = 120 kWpf1 = 0.75pf2 = 0.9VL = 480 Vf = 60 Hz
Cph = µF
3.TransformersMedium
A single-phase 250 kVA, 6600/230 V transformer operates at its rated load. Treat it as ideal. Find the current IHV in the high-voltage winding.
S = 250 kVAV1 = 6600 VV2 = 230 V
IHV = A
4.TransformersHard
A single-phase 25 kVA transformer has a core loss of 200 W. At rated load and a power factor of 0.8 lagging its efficiency is 96%. Find the full-load copper loss Pcu,fl in watts.
S = 25 kVAPcore = 200 Wpf = 0.8η = 96%
Pcu,fl = W
5.AC power & power factorMedium
A single-phase motor takes 2 kW at a power factor of 0.7 lagging from a 240 V supply. Find the reactive power Q it draws.
P = 2 kWpf = 0.7V = 240 V
Q = kvar
6.AC power & power factorHard
A 200 kVA single-phase transformer supplies a load of 83 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 = 200 kVAP1 = 83 kWpf1 = 0.65
ΔP = kW
7.AC impedance & phasorsMedium
A series RLC circuit is connected to a 10 V (rms), 500 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).
V = 10 V (rms)f = 500 HzR = 30 ΩL = 250 mHC = 1 µF
φ = °
8.AC impedance & phasorsHard
A series RLC circuit has R = 5 Ω, L = 5 mH and C = 0.047 µF. Find its quality factor Q at resonance.
R = 5 ΩL = 5 mHC = 0.047 µF
Q =
9.DC motorsMedium
A DC motor delivers 5.5 kW of mechanical power at 750 rpm. Find the shaft torque T.
P = 5.5 kWn = 750 rpm
T = N·m
10.DC motorsHard
A 120 V DC motor has an armature resistance of 0.25 Ω. Find the series starting resistance Rst that limits the starting current to 40 A.
V = 120 VRa = 0.25 ΩImax = 40 A
Rst = Ω
3Solutions
1.Three-phase power
Answer: P = 12.29 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 = 480/√3 = 277.1 V
Phase current
|Z| = √(12² + 9²) = 15 Ω
Iph = 277.1/15 = 18.48 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·18.48²·12 = 12.29 kW
Answer
P = 12.29 kW
Check: cos φ = R/|Z| = 0.8, and √3·VL·IL·cos φ gives the same power.
2.Three-phase power
Answer: Cph = 549.3 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 = 120·tan(acos 0.75) = 120·0.8819 = 105.8 kvar
Q2 = 120·tan(acos 0.9) = 120·0.4843 = 58.12 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 47.71 kvar
Capacitance per phase
In a Y bank each capacitor is connected between a line and the star point.
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.
IHV = S/V = 250000/6600 = 37.88 A
Answer
IHV = 37.88 A
4.Transformers
Answer: Pcu,fl = 633.3 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 = 25·0.8 = 20 kW
Pin = 20000/0.96 = 20833 W
Total and copper losses
Ploss = 20833 − 20000 = 833.3 W
Pcu,fl = 833.3 − 200 = 633.3 W
Answer
Pcu,fl = 633.3 W
5.AC power & power factor
Answer: Q = 2.04 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 = 2·1.02 = 2.04 kvar
Answer
Q = 2.04 kvar
6.AC power & power factor
Answer: ΔP = 91.88 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 = 83·tan(arccos 0.65) = 97.04 kvar
Largest total real power
At full rating S = Smax, and Q is unchanged by the resistive load.
P = √(Smax² − Q1²) = √(200² − 97.04²) = 174.9 kW
Extra real power
ΔP = 174.9 − 83 = 91.88 kW
Answer
ΔP = 91.88 kW
7.AC impedance & phasors
Answer: φ = 86.33 °
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 = 3142·0.25 = 785.4 Ω
XC = 1/(ωC) = 1/(3142·1·10⁻⁶) = 318.3 Ω
Impedance
In series the impedances add as complex numbers: Z = R + j(XL − XC).
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·√(0.005·0.047·10⁻⁶)) = 10382 Hz
Quality factor
Q compares the reactance at resonance with the resistance: Q = ω₀L/R.
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·750/60 = 78.54 rad/s
Torque
T = P/ω = 5500/78.54 = 70.03 N·m
Answer
T = 70.03 N·m
10.DC motors
Answer: Rst = 2.75 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.
Total resistance
Ra + Rst = 120/40 = 3 Ω
Starting resistor
Rst = 3 − 0.25 = 2.75 Ω
Answer
Rst = 2.75 Ω
The resistor is cut out step by step as the motor speeds up and the back-EMF grows.