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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 Δ-connected load with an impedance of 18 + j24 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 400 V. Find the total real power P absorbed by the load in kW.
VL = 400 VZ = 18 + j24 Ω per phase
P = kW
2.Three-phase powerHard
A balanced three-phase load takes 60 kW at a power factor of 0.78 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 0.92 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.
P = 60 kWpf1 = 0.78pf2 = 0.92VL = 415 Vf = 60 Hz
Cph = µF
3.TransformersMedium
A single-phase 50 kVA, 2400/240 V transformer operates at its rated load. Treat it as ideal. Find the current ILV in the low-voltage winding.
S = 50 kVAV1 = 2400 VV2 = 240 V
ILV = A
4.TransformersHard
A single-phase 250 kVA transformer has a core loss of 1000 W. At rated load and a power factor of 0.8 lagging its efficiency is 99%. Find the full-load copper loss Pcu,fl in watts.
S = 250 kVAPcore = 1000 Wpf = 0.8η = 99%
Pcu,fl = W
5.AC power & power factorMedium
A load with impedance Z = 7 + j24 Ω is connected to a 120 V (rms) single-phase supply. Find the reactive power Q it absorbs.
V = 120 VZ = 7 + j24 Ω
Q = var
6.AC power & power factorHard
A single-phase load takes 25 kW at a power factor of 0.65 lagging from a 230 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the new line current Inew.
P = 25 kWpf1 = 0.65pf2 = 0.98V = 230 Vf = 60 Hz
Inew = A
7.AC impedance & phasorsMedium
A series RLC circuit with L = 100 mH must resonate at 1000 Hz. Find the capacitance C needed.
f0 = 1000 HzL = 100 mH
C = µF
8.AC impedance & phasorsHard
A series RLC circuit is connected to a 24 V (rms), 1000 Hz supply. Find the rms voltage VC across the capacitor.
V = 24 V (rms)f = 1000 HzR = 10 ΩL = 20 mHC = 100 µF
VC = V
9.DC motorsMedium
A DC motor delivers 0.5 kW of mechanical power at 1450 rpm. Find the shaft torque T.
P = 0.5 kWn = 1450 rpm
T = N·m
10.DC motorsHard
A DC motor (k = 0.8 V·s/rad, Ra = 0.25 Ω) on a fixed 110 V supply drives a load of 12 N·m. Find the series resistance Rs that slows it down to 600 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.
Phase voltage
Vph = VL = 400 V
Phase current
|Z| = √(18² + 24²) = 30 Ω
Iph = 400/30 = 13.33 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·13.33²·18 = 9.6 kW
Answer
P = 9.6 kW
Check: cos φ = R/|Z| = 0.6, and √3·VL·IL·cos φ gives the same power.
2.Three-phase power
Answer: Cph = 115.9 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 = 60·tan(acos 0.78) = 60·0.8023 = 48.14 kvar
Q2 = 60·tan(acos 0.92) = 60·0.426 = 25.56 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 22.58 kvar
Capacitance per phase
In a Δ bank each capacitor is connected between two lines.
V = 415 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 7526/(377·415²) = 115.9 µF
Answer
Cph = 115.9 µF
3.Transformers
Answer: ILV = 208.3 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.
Rated current
Both windings carry the same apparent power.
ILV = S/V = 50000/240 = 208.3 A
Answer
ILV = 208.3 A
4.Transformers
Answer: Pcu,fl = 1020 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 = 250·0.8 = 200 kW
Pin = 200000/0.99 = 202020 W
Total and copper losses
Ploss = 202020 − 200000 = 2020 W
Pcu,fl = 2020 − 1000 = 1020 W
Answer
Pcu,fl = 1020 W
5.AC power & power factor
Answer: Q = 553 var
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.
Current
|Z| = √(7² + 24²) = 25 Ω
I = V/|Z| = 120/25 = 4.8 A
Reactive power
Q = I²·X = 4.8²·24 = 553 var
Answer
Q = 553 var
6.AC power & power factor
Answer: Inew = 110.9 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 = P·tan φ1 = 25·tan(49.46°) = 29.23 kvar
Q2 = P·tan φ2 = 25·tan(11.48°) = 5.076 kvar
QC = Q1 − Q2 = 24.15 kvar
New line current
Inew = P/(V·cos φ2) = 25000/(230·0.98) = 110.9 A
Answer
Inew = 110.9 A
7.AC impedance & phasors
Answer: C = 0.2533 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·1000 = 6283 rad/s
C = 1/(ω₀²·L) = 1/(6283²·0.1) = 0.2533 µF
Answer
C = 0.2533 µF
8.AC impedance & phasors
Answer: VC = 0.3069 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|.
Reactances
ω = 2πf = 2π·1000 = 6283 rad/s
XL = ωL = 6283·0.02 = 125.7 Ω
XC = 1/(ωC) = 1/(6283·100·10⁻⁶) = 1.592 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| = 24/124.5 = 0.1928 A
Voltage across the capacitor
VC = I·XC = 0.1928·1.592 = 0.3069 V
Answer
VC = 0.3069 V
9.DC motors
Answer: T = 3.293 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·1450/60 = 151.8 rad/s
Torque
T = P/ω = 500/151.8 = 3.293 N·m
Answer
T = 3.293 N·m
10.DC motors
Answer: Rs = 3.732 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 = 12/0.8 = 15 A
E = 0.8·62.83 = 50.27 V
Series resistance
Ra + Rs = (110 − 50.27)/15 = 3.982 Ω
Rs = 3.732 Ω
Answer
Rs = 3.732 Ω
The resistor wastes 839.8 W; a chopper controls the speed with far less loss.