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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 10 A from a supply with a line-to-line voltage of 415 V. The power factor is 0.82 lagging. Find the total real power P drawn by the load in kW.
VL = 415 VIL = 10 Acos φ = 0.82
P = kW
2.Three-phase powerHard
A balanced three-phase load takes 60 kW at a power factor of 0.8 lagging from a 415 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 = 60 kWpf1 = 0.8pf2 = 0.9VL = 415 Vf = 60 Hz
Cph = µF
3.TransformersMedium
An ideal transformer with a turns ratio N₁:N₂ = 10:1 has 2400 V applied to its primary winding. A 30 Ω resistor is connected across the secondary winding. Find the primary current I1.
N1:N2 = 10:1V1 = 2400 VRL = 30 Ω
I1 = A
4.TransformersHard
A single-phase 150 kVA transformer has a core loss of 1200 W. At rated load and a power factor of 0.9 lagging its efficiency is 96%. Find the full-load copper loss Pcu,fl in watts.
S = 150 kVAPcore = 1200 Wpf = 0.9η = 96%
Pcu,fl = W
5.AC power & power factorMedium
Two loads are connected in parallel to a 240 V single-phase supply: a 19 kW load at a power factor of 0.85 lagging and a 20 kW load at a power factor of 0.95 leading. Find the overall power factor cos φ seen by the supply.
V = 240 VP1 = 19 kW, pf1 = 0.85 laggingP2 = 20 kW, pf2 = 0.95 leading
cos φ =
6.AC power & power factorHard
A single-phase load takes 30 kW at a power factor of 0.8 lagging from a 240 V, 50 Hz supply. A 470 µF capacitor is connected in parallel with the load. Find the new line current I.
P = 30 kWpf1 = 0.8V = 240 Vf = 50 HzC = 470 µF
I = A
7.AC impedance & phasorsMedium
A series RLC circuit is connected to a 120 V (rms), 60 Hz supply. Find the rms current I.
V = 120 V (rms)f = 60 HzR = 10 ΩL = 50 mHC = 22 µF
I = A
8.AC impedance & phasorsHard
A resistor R1 is connected in series with a parallel combination of R2 and a capacitor C, across a 24 V (rms), 500 Hz supply. Find the rms current I.
V = 24 V (rms)f = 500 HzR1 = 20 ΩR2 = 30 ΩC = 4.7 µF
I = A
9.DC motorsMedium
A DC motor with k = 2 N·m/A and Ra = 1.2 Ω is connected to 220 V. Find its stall torque Tstall, the torque at zero speed.
V = 220 Vk = 2 N·m/ARa = 1.2 Ω
Tstall = N·m
10.DC motorsHard
A small DC motor has an armature resistance of 3 Ω and runs from 110 V. Neglecting rotational losses, find the largest mechanical power Pmax it can develop at any speed.
V = 110 VRa = 3 Ω
Pmax = W
3Solutions
1.Three-phase power
Answer: P = 5.894 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 = √3·415·10 = 7.188 kVA
Real power
P = S·cos φ = 7.188·0.82 = 5.894 kW
Answer
P = 5.894 kW
2.Three-phase power
Answer: Cph = 245.5 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 = 60·tan(acos 0.8) = 60·0.75 = 45 kvar
Q2 = 60·tan(acos 0.9) = 60·0.4843 = 29.06 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 15.94 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.
Secondary voltage and current
V2 = 2400/10 = 240 V
I2 = 240/30 = 8 A
Primary current
An ideal transformer has no losses: V₁·I₁ = V₂·I₂.
I1 = I2/10 = 0.8 A
Answer
I1 = 0.8 A
Check: the load seen from the primary is a²·RL = 3000 Ω, and 2400/3000 gives the same current.
4.Transformers
Answer: Pcu,fl = 4425 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 = 150·0.9 = 135 kW
Pin = 135000/0.96 = 140625 W
Total and copper losses
Ploss = 140625 − 135000 = 5625 W
Pcu,fl = 5625 − 1200 = 4425 W
Answer
Pcu,fl = 4425 W
5.AC power & power factor
Answer: cos φ = 0.9912
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 each load
A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).
Q1 = 19·tan(arccos 0.85) = 11.78 kvar
Q2 = −20·tan(arccos 0.95) = −6.574 kvar
Totals
P = 19 + 20 = 39 kW
Q = 11.78 + (−6.574) = 5.201 kvar
S = √(39² + 5.201²) = 39.35 kVA
Overall power factor
cos φ = P/S = 39/39.35 = 0.9912 lagging
Answer
cos φ = 0.9912
6.AC power & power factor
Answer: I = 137.9 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 = 30·tan(arccos 0.8) = 22.5 kvar
QC = V²·2πf·C = 240²·2π·50·470·10⁻⁶ = 8505 var = 8.505 kvar
Q2 = 22.5 − 8.505 = 14 kvar
New apparent power
S2 = √(30² + 14²) = 33.1 kVA
New line current
I = S2/V = 33104/240 = 137.9 A
Answer
I = 137.9 A
7.AC impedance & phasors
Answer: I = 1.174 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·60 = 377 rad/s
XL = ωL = 377·0.05 = 18.85 Ω
XC = 1/(ωC) = 1/(377·22·10⁻⁶) = 120.6 Ω
Impedance
In series the impedances add as complex numbers: Z = R + j(XL − XC).
The same current flows through every element of a series circuit.
I = V/|Z| = 120/102.2 = 1.174 A
Answer
I = 1.174 A
8.AC impedance & phasors
Answer: I = 0.5169 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|.
Reactance
XC = 1/(2π·500·4.7·10⁻⁶) = 67.73 Ω
Parallel combination
Multiply top and bottom by the complex conjugate to split Zp into real and imaginary parts.
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.
Stall current
Ia = V/Ra = 220/1.2 = 183.3 A
Stall torque
Tstall = 2·183.3 = 366.7 N·m
Answer
Tstall = 366.7 N·m
10.DC motors
Answer: Pmax = 1008 W
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.
Power as a function of the back-EMF
P(E) = E·(V − E)/Ra
dP/dE = 0 → E = V/2 = 55 V
Maximum power
Pmax = (V/2)²/Ra = V²/(4Ra) = 110²/(4·3) = 1008 W
Answer
Pmax = 1008 W
At this point the efficiency is only 50 %, so real motors run well above half speed.