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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 three-phase induction motor delivers 22 kW of mechanical output power at full load. At this load its efficiency is 93% and its power factor is 0.86 lagging. The supply has a line-to-line voltage of 400 V. Find the line current IL.
Pout = 22 kWη = 93%cos φ = 0.86VL = 400 V
IL = A
2.Three-phase powerHard
A balanced three-phase load takes 150 kW at a power factor of 0.65 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 0.95 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.
P = 150 kWpf1 = 0.65pf2 = 0.95VL = 400 Vf = 50 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 75 W. At rated load and a power factor of 0.8 lagging its efficiency is 98.2%. Find the full-load copper loss Pcu,fl in watts.
S = 25 kVAPcore = 75 Wpf = 0.8η = 98.2%
Pcu,fl = W
5.AC power & power factorMedium
An inductive load takes 6.5 A and 897 W from a 230 V (rms) single-phase supply. Model it as a resistance R in series with a reactance X. Find R.
V = 230 VI = 6.5 AP = 897 W
R = Ω
6.AC power & power factorHard
A single-phase load takes 29 kW at a power factor of 0.6 lagging from a 230 V, 60 Hz supply. A 220 µF capacitor is connected in parallel with the load. Find the new power factor cos φ seen by the supply.
P = 29 kWpf1 = 0.6V = 230 Vf = 60 HzC = 220 µF
cos φ =
7.AC impedance & phasorsMedium
A series RLC circuit with L = 5 mH must resonate at 100 Hz. Find the capacitance C needed.
f0 = 100 HzL = 5 mH
C = µF
8.AC impedance & phasorsHard
A series RLC circuit is connected to a 24 V (rms), 500 Hz supply. Find the resonant frequency f0 of the circuit.
V = 24 V (rms)f = 500 HzR = 15 ΩL = 100 mHC = 1 µF
f0 = Hz
9.DC motorsMedium
A DC motor delivers 7.5 kW of mechanical power at 1800 rpm. Find the shaft torque T.
P = 7.5 kWn = 1800 rpm
T = N·m
10.DC motorsHard
A separately excited DC motor on 220 V (Ra = 2 Ω) runs at 1200 rpm with an armature current of 4 A. The load changes and the current becomes 6 A, with the field unchanged. Find the new speed n2.
V = 220 VRa = 2 Ωn1 = 1200 rpmIa1 = 4 AIa2 = 6 A
n2 = rpm
3Solutions
1.Three-phase power
Answer: IL = 39.7 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 = 22/0.93 = 23.66 kW
Line current
IL = Pin/(√3·VL·cos φ) = 23656/(√3·400·0.86) = 39.7 A
Answer
IL = 39.7 A
2.Three-phase power
Answer: Cph = 2508 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 = 150·tan(acos 0.65) = 150·1.169 = 175.4 kvar
Q2 = 150·tan(acos 0.95) = 150·0.3287 = 49.3 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 126.1 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 = 291.6 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 = 25·0.8 = 20 kW
Pin = 20000/0.982 = 20367 W
Total and copper losses
Ploss = 20367 − 20000 = 366.6 W
Pcu,fl = 366.6 − 75 = 291.6 W
Answer
Pcu,fl = 291.6 W
5.AC power & power factor
Answer: R = 21.23 Ω
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.
Resistance from the real power
R = P/I² = 897/6.5² = 21.23 Ω
Answer
R = 21.23 Ω
6.AC power & power factor
Answer: cos φ = 0.6459
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 = 29·tan(arccos 0.6) = 38.67 kvar
QC = V²·2πf·C = 230²·2π·60·220·10⁻⁶ = 4387 var = 4.387 kvar
Q2 = 38.67 − 4.387 = 34.28 kvar
New apparent power
S2 = √(29² + 34.28²) = 44.9 kVA
New power factor
cos φ = P/S2 = 29/44.9 = 0.6459
Answer
cos φ = 0.6459
7.AC impedance & phasors
Answer: C = 506.6 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·100 = 628.3 rad/s
C = 1/(ω₀²·L) = 1/(628.3²·0.005) = 506.6 µF
Answer
C = 506.6 µF
8.AC impedance & phasors
Answer: f0 = 503.3 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.
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·1800/60 = 188.5 rad/s
Torque
T = P/ω = 7500/188.5 = 39.79 N·m
Answer
T = 39.79 N·m
10.DC motors
Answer: n2 = 1177 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.