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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 three-phase load draws a line current of 12 A from a supply with a line-to-line voltage of 208 V. The power factor is 0.95 lagging. Find the total real power P drawn by the load in kW.
VL = 208 VIL = 12 Acos φ = 0.95
P = kW
2.Three-phase powerHard
A balanced three-phase load takes 80 kW at a power factor of 0.65 lagging from a 480 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 = 80 kWpf1 = 0.65pf2 = 0.95VL = 480 Vf = 50 Hz
Cph = µF
3.TransformersMedium
A single-phase 50 kVA, 6600/230 V transformer operates at its rated load. Treat it as ideal. Find the current IHV in the high-voltage winding.
S = 50 kVAV1 = 6600 VV2 = 230 V
IHV = A
4.TransformersHard
A single-phase 75 kVA transformer has a core loss of 300 W and a copper loss of 600 W at rated load. It supplies 50% of its rated kVA at a power factor of 0.8 lagging. Find the total losses Ploss in watts at this load.
A single-phase motor takes 15 kW at a power factor of 0.6 lagging from a 400 V supply. Find the reactive power Q it draws.
P = 15 kWpf = 0.6V = 400 V
Q = kvar
6.AC power & power factorHard
A single-phase load takes 2 kW at a power factor of 0.65 lagging from a 400 V, 60 Hz supply. A 33 µF capacitor is connected in parallel with the load. Find the new line current I.
P = 2 kWpf1 = 0.65V = 400 Vf = 60 HzC = 33 µF
I = A
7.AC impedance & phasorsMedium
A series RLC circuit with C = 10 µF must resonate at 200 Hz. Find the inductance L needed.
f0 = 200 HzC = 10 µF
L = mH
8.AC impedance & phasorsHard
A series RLC circuit is driven by a 5 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.
V = 5 V (rms)R = 5 ΩL = 10 mHC = 2.2 µF
VL = V
9.DC motorsMedium
A DC motor (k = 0.6 V·s/rad, Ra = 2 Ω) runs from 240 V. At full load it draws 10 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.
V = 240 Vk = 0.6 V·s/radRa = 2 ΩIFL = 10 A
SR = %
10.DC motorsHard
A DC motor (k = 0.2 V·s/rad, Ra = 0.5 Ω) on a fixed 24 V supply drives a load of 1 N·m. Find the series resistance Rs that slows it down to 800 rpm.
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 = √3·208·12 = 4.323 kVA
Real power
P = S·cos φ = 4.323·0.95 = 4.107 kW
Answer
P = 4.107 kW
2.Three-phase power
Answer: Cph = 928.9 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 = 80·tan(acos 0.65) = 80·1.169 = 93.53 kvar
Q2 = 80·tan(acos 0.95) = 80·0.3287 = 26.29 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 67.24 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 = 50000/6600 = 7.576 A
Answer
IHV = 7.576 A
4.Transformers
Answer: Ploss = 450 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.
Copper loss at this load
Pcu = x²·Pcu,fl = 0.25·600 = 150 W
Total losses
The core loss is the same at any load.
Ploss = 300 + 150 = 450 W
Answer
Ploss = 450 W
The efficiency at this load is η = 98.52%.
5.AC power & power factor
Answer: Q = 20 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(0.6) = 53.13°, tan φ = 1.333
Reactive power
Q = P·tan φ = 15·1.333 = 20 kvar
Answer
Q = 20 kvar
6.AC power & power factor
Answer: I = 5.075 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 = 2·tan(arccos 0.65) = 2.338 kvar
QC = V²·2πf·C = 400²·2π·60·33·10⁻⁶ = 1991 var = 1.991 kvar
Q2 = 2.338 − 1.991 = 0.3477 kvar
New apparent power
S2 = √(2² + 0.3477²) = 2.03 kVA
New line current
I = S2/V = 2030/400 = 5.075 A
Answer
I = 5.075 A
7.AC impedance & phasors
Answer: L = 63.33 mH
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·200 = 1257 rad/s
L = 1/(ω₀²·C) = 1/(1257²·10·10⁻⁶) = 63.33 mH
Answer
L = 63.33 mH
8.AC impedance & phasors
Answer: VL = 67.42 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/√(0.01·2.2·10⁻⁶) = 6742 rad/s
I = V/R = 5/5 = 1 A
Voltage across the inductor
XL = ω₀L = 6742·0.01 = 67.42 Ω
VL = I·XL = 1·67.42 = 67.42 V
Answer
VL = 67.42 V
This is Q = 13.48 times the supply voltage.
9.DC motors
Answer: SR = 9.091 %
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.
No-load and full-load speed
n₀ = 60·240/(2π·0.6) = 3820 rpm
nFL = 60·(240 − 10·2)/(2π·0.6) = 3501 rpm
Speed regulation
SR = (3820 − 3501)/3501·100 = 9.091 %
Answer
SR = 9.091 %
10.DC motors
Answer: Rs = 0.949 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 = 1/0.2 = 5 A
E = 0.2·83.78 = 16.76 V
Series resistance
Ra + Rs = (24 − 16.76)/5 = 1.449 Ω
Rs = 0.949 Ω
Answer
Rs = 0.949 Ω
The resistor wastes 23.72 W; a chopper controls the speed with far less loss.