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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 30 A from a supply with a line-to-line voltage of 690 V. The power factor is 0.7 lagging. Find the total real power P drawn by the load in kW.
VL = 690 VIL = 30 Acos φ = 0.7
P = kW
2.Three-phase powerHard
A balanced three-phase load takes 75 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 0.98 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.
P = 75 kWpf1 = 0.78pf2 = 0.98VL = 415 Vf = 60 Hz
QC = kvar
3.TransformersMedium
An ideal transformer with a turns ratio N₁:N₂ = 4:1 has 960 V applied to its primary winding. A 12 Ω resistor is connected across the secondary winding. Find the primary current I1.
N1:N2 = 4:1V1 = 960 VRL = 12 Ω
I1 = A
4.TransformersHard
A single-phase 25 kVA transformer has a core loss of 100 W and a copper loss of 250 W at rated load. It supplies 75% 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 12 kW at a power factor of 0.8 lagging from a 230 V supply. Find the reactive power Q it draws.
P = 12 kWpf = 0.8V = 230 V
Q = kvar
6.AC power & power factorHard
A single-phase load takes 20 kW at a power factor of 0.7 lagging from a 400 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the reactive power QC the capacitor must supply.
P = 20 kWpf1 = 0.7pf2 = 0.98V = 400 Vf = 50 Hz
QC = kvar
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), 500 Hz supply. Find the resonant frequency f0 of the circuit.
V = 24 V (rms)f = 500 HzR = 30 ΩL = 20 mHC = 10 µF
f0 = Hz
9.DC motorsMedium
A DC motor (k = 0.75 N·m/A, Ra = 0.8 Ω) runs from 220 V and drives a load torque of 18.75 N·m. Find its speed n in rpm.
V = 220 Vk = 0.75 N·m/ARa = 0.8 ΩT = 18.75 N·m
n = rpm
10.DC motorsHard
A small DC motor has an armature resistance of 1 Ω and runs from 24 V. Neglecting rotational losses, find the largest mechanical power Pmax it can develop at any speed.
V = 24 VRa = 1 Ω
Pmax = W
3Solutions
1.Three-phase power
Answer: P = 25.1 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 = √3·690·30 = 35.85 kVA
Real power
P = S·cos φ = 35.85·0.7 = 25.1 kW
Answer
P = 25.1 kW
2.Three-phase power
Answer: QC = 44.94 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 = 75·tan(acos 0.78) = 75·0.8023 = 60.17 kvar
Q2 = 75·tan(acos 0.98) = 75·0.2031 = 15.23 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 44.94 kvar
Answer
QC = 44.94 kvar
3.Transformers
Answer: I1 = 5 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 = 960/4 = 240 V
I2 = 240/12 = 20 A
Primary current
An ideal transformer has no losses: V₁·I₁ = V₂·I₂.
I1 = I2/4 = 5 A
Answer
I1 = 5 A
Check: the load seen from the primary is a²·RL = 192 Ω, and 960/192 gives the same current.
4.Transformers
Answer: Ploss = 240.6 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.
Copper loss at this load
Pcu = x²·Pcu,fl = 0.5625·250 = 140.6 W
Total losses
The core loss is the same at any load.
Ploss = 100 + 140.6 = 240.6 W
Answer
Ploss = 240.6 W
The efficiency at this load is η = 98.42%.
5.AC power & power factor
Answer: Q = 9 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(0.8) = 36.87°, tan φ = 0.75
Reactive power
Q = P·tan φ = 12·0.75 = 9 kvar
Answer
Q = 9 kvar
6.AC power & power factor
Answer: QC = 16.34 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.
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 = 20·tan(45.57°) = 20.4 kvar
Q2 = P·tan φ2 = 20·tan(11.48°) = 4.061 kvar
QC = Q1 − Q2 = 16.34 kvar
Answer
QC = 16.34 kvar
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: f0 = 355.9 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.
Armature current
Ia = T/k = 18.75/0.75 = 25 A
Back-EMF and speed
E = 220 − 25·0.8 = 200 V
ω = E/k = 266.7 rad/s → n = 2546 rpm
Answer
n = 2546 rpm
10.DC motors
Answer: Pmax = 144 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 = 12 V
Maximum power
Pmax = (V/2)²/Ra = V²/(4Ra) = 24²/(4·1) = 144 W
Answer
Pmax = 144 W
At this point the efficiency is only 50 %, so real motors run well above half speed.