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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 45 kW of mechanical output power at full load. At this load its efficiency is 92% 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 = 45 kWη = 92%cos φ = 0.86VL = 400 V
IL = A
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. Find the total reactive power QC the capacitor bank must supply, in kvar.
P = 80 kWpf1 = 0.65pf2 = 0.95VL = 480 Vf = 50 Hz
QC = kvar
3.TransformersMedium
An ideal transformer with a turns ratio N₁:N₂ = 2:1 has 480 V applied to its primary winding. A 6 Ω resistor is connected across the secondary winding. Find the primary current I1.
N1:N2 = 2:1V1 = 480 VRL = 6 Ω
I1 = A
4.TransformersHard
A single-phase 200 kVA transformer has a core loss of 1000 W and a copper loss of 1500 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?
S = 200 kVAPcore = 1000 WPcu,fl = 1500 W
Sη,max = kVA
5.AC power & power factorMedium
A single-phase motor takes 6 kW at a power factor of 0.9 lagging from a 240 V supply. Find the reactive power Q it draws.
P = 6 kWpf = 0.9V = 240 V
Q = kvar
6.AC power & power factorHard
A 120 V (rms), 60 Hz source has an internal impedance Zs = 8 + j40 Ω. The load impedance ZL can be chosen freely. The load for maximum power is built from a resistor in series with a capacitor. Find the capacitance C.
Vs = 120 V (rms)f = 60 HzZs = 8 + j40 Ω
C = µF
7.AC impedance & phasorsMedium
A series RLC circuit with C = 4.7 µF must resonate at 500 Hz. Find the inductance L needed.
f0 = 500 HzC = 4.7 µF
L = mH
8.AC impedance & phasorsHard
A series RLC circuit is driven by a 10 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.
V = 10 V (rms)R = 10 ΩL = 10 mHC = 0.47 µF
VL = V
9.DC motorsMedium
A DC motor delivers 0.5 kW of mechanical power at 1000 rpm. Find the shaft torque T.
P = 0.5 kWn = 1000 rpm
T = N·m
10.DC motorsHard
A separately excited DC motor on 120 V (Ra = 1.5 Ω) runs at 1440 rpm with an armature current of 6 A. The load changes and the current becomes 12 A, with the field unchanged. Find the new speed n2.
V = 120 VRa = 1.5 Ωn1 = 1440 rpmIa1 = 6 AIa2 = 12 A
n2 = rpm
3Solutions
1.Three-phase power
Answer: IL = 82.09 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 = 45/0.92 = 48.91 kW
Line current
IL = Pin/(√3·VL·cos φ) = 48913/(√3·400·0.86) = 82.09 A
Answer
IL = 82.09 A
2.Three-phase power
Answer: QC = 67.24 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 = 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
Answer
QC = 67.24 kvar
3.Transformers
Answer: I1 = 20 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.
Secondary voltage and current
V2 = 480/2 = 240 V
I2 = 240/6 = 40 A
Primary current
An ideal transformer has no losses: V₁·I₁ = V₂·I₂.
I1 = I2/2 = 20 A
Answer
I1 = 20 A
Check: the load seen from the primary is a²·RL = 24 Ω, and 480/24 gives the same current.
4.Transformers
Answer: Sη,max = 163.3 kVA
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 equals core loss
x²·1500 = 1000 → x = √(1000/1500) = 0.8165
Load at maximum efficiency
Sη,max = x·S = 0.8165·200 = 163.3 kVA
Answer
Sη,max = 163.3 kVA
5.AC power & power factor
Answer: Q = 2.906 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(0.9) = 25.84°, tan φ = 0.4843
Reactive power
Q = P·tan φ = 6·0.4843 = 2.906 kvar
Answer
Q = 2.906 kvar
6.AC power & power factor
Answer: C = 66.31 µF
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* = 8 − j40 Ω
Capacitance
XC = 40 Ω → C = 1/(2π·60·40) = 66.31 µF
Answer
C = 66.31 µF
7.AC impedance & phasors
Answer: L = 21.56 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·500 = 3142 rad/s
L = 1/(ω₀²·C) = 1/(3142²·4.7·10⁻⁶) = 21.56 mH
Answer
L = 21.56 mH
8.AC impedance & phasors
Answer: VL = 145.9 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/√(0.01·0.47·10⁻⁶) = 14586 rad/s
I = V/R = 10/10 = 1 A
Voltage across the inductor
XL = ω₀L = 14586·0.01 = 145.9 Ω
VL = I·XL = 1·145.9 = 145.9 V
Answer
VL = 145.9 V
This is Q = 14.59 times the supply voltage.
9.DC motors
Answer: T = 4.775 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·1000/60 = 104.7 rad/s
Torque
T = P/ω = 500/104.7 = 4.775 N·m
Answer
T = 4.775 N·m
10.DC motors
Answer: n2 = 1323 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.