An ideal boost converter turns 12 V into 20 V for a 4 Ω load at 250 kHz. Find the smallest inductance Lcrit for continuous conduction.
Vin = 12 VVo = 20 Vf = 250 kHzR = 4 Ω
Lcrit = µH
3.DC motorsMedium
A DC motor with k = 0.6 V·s/rad runs from 120 V. Neglecting all losses at no load, find its no-load speed n0 in rpm.
V = 120 Vk = 0.6 V·s/rad
n0 = rpm
4.DC motorsHard
A separately excited DC motor on 240 V (Ra = 1 Ω) runs at 900 rpm with an armature current of 12 A. The load changes and the current becomes 6 A, with the field unchanged. Find the new speed n2.
V = 240 VRa = 1 Ωn1 = 900 rpmIa1 = 12 AIa2 = 6 A
n2 = rpm
5.AC power & power factorMedium
A single-phase motor takes 8 kW at a power factor of 0.6 lagging from a 230 V supply. Find the line current I.
P = 8 kWpf = 0.6V = 230 V
I = A
6.AC power & power factorHard
A 230 V (rms), 1000 Hz source has an internal impedance Zs = 4 + j20 Ω. The load impedance ZL can be chosen freely. Find the maximum real power Pmax that can be delivered to the load.
Vs = 230 V (rms)f = 1000 HzZs = 4 + j20 Ω
Pmax = W
7.Three-phase powerMedium
A balanced Y-connected load with an impedance of 12 + j5 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 415 V. Find the total real power P absorbed by the load in kW.
VL = 415 VZ = 12 + j5 Ω per phase
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 200 kW at a power factor of 0.8 lagging from a 690 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.98 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.
P = 200 kWpf1 = 0.8pf2 = 0.98VL = 690 Vf = 50 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: Imax = 0.3132 A
DC-DC converters: a switch S turns on for the fraction D of each period T = 1/f. In steady state the average inductor voltage is zero (volt-second balance), which gives the ideal conversion ratios in continuous conduction: buck Vo = D·Vin, boost Vo = Vin/(1 − D), inverting buck-boost |Vo| = Vin·D/(1 − D). An ideal converter keeps the power: Vin·Iin = Vo·Io. While S is on the inductor current rises by ΔIL = VL·D·T/L, with VL = Vin − Vo for the buck and Vin for the boost and buck-boost. The output ripple is ΔVo = ΔIL/(8fC) for the buck and Io·D/(fC) for the boost and buck-boost. The inductor current just touches zero (the CCM boundary) when the average inductor current equals ΔIL/2.
Duty cycle
D = 5.4/9 = 0.6
Average and ripple current
Io = 5.4/20 = 0.27 A
ΔIL = VL·D/(L·f) = 3.6·0.6/(100·10⁻⁶·250·10³) = 0.0864 A
Peak current
Imax = 0.27 + 0.0864/2 = 0.3132 A
Answer
Imax = 0.3132 A
2.DC-DC converters
Answer: Lcrit = 1.152 µH
DC-DC converters: a switch S turns on for the fraction D of each period T = 1/f. In steady state the average inductor voltage is zero (volt-second balance), which gives the ideal conversion ratios in continuous conduction: buck Vo = D·Vin, boost Vo = Vin/(1 − D), inverting buck-boost |Vo| = Vin·D/(1 − D). An ideal converter keeps the power: Vin·Iin = Vo·Io. While S is on the inductor current rises by ΔIL = VL·D·T/L, with VL = Vin − Vo for the buck and Vin for the boost and buck-boost. The output ripple is ΔVo = ΔIL/(8fC) for the buck and Io·D/(fC) for the boost and buck-boost. The inductor current just touches zero (the CCM boundary) when the average inductor current equals ΔIL/2.
Duty cycle
D = 1 − 12/20 = 0.4
Boundary condition
Set the average inductor current equal to half the ripple current and solve for L.
Lcrit = D(1 − D)²·R/(2f)
Critical inductance
Lcrit = 0.4(1 − 0.4)²·4/(2250·10³) = 1.152 µH
Answer
Lcrit = 1.152 µH
3.DC motors
Answer: n0 = 1910 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.
No-load speed
ω₀ = V/k = 120/0.6 = 200 rad/s
n0 = 60·200/(2π) = 1910 rpm
Answer
n0 = 1910 rpm
4.DC motors
Answer: n2 = 923.7 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.
Back-EMF before and after
E1 = 240 − 12·1 = 228 V
E2 = 240 − 6·1 = 234 V
New speed
n2 = 900·234/228 = 923.7 rpm
Answer
n2 = 923.7 rpm
5.AC power & power factor
Answer: I = 57.97 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.
Line current
Only the real part of the power does work, so divide by the power factor.
I = P/(V·cos φ) = 8000/(230·0.6) = 57.97 A
Answer
I = 57.97 A
6.AC power & power factor
Answer: Pmax = 3306 W
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* = 4 − j20 Ω
Maximum power
I = Vs/(2Rs) = 230/8 = 28.75 A
Pmax = I²·RL = Vs²/(4Rs) = 230²/(4·4) = 3306 W
Answer
Pmax = 3306 W
7.Three-phase power
Answer: P = 12.23 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.
Phase voltage
Vph = 415/√3 = 239.6 V
Phase current
|Z| = √(12² + 5²) = 13 Ω
Iph = 239.6/13 = 18.43 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·18.43²·12 = 12.23 kW
Answer
P = 12.23 kW
Check: cos φ = R/|Z| = 0.9231, and √3·VL·IL·cos φ gives the same power.
8.Three-phase power
Answer: Cph = 243.8 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 = 200·tan(acos 0.8) = 200·0.75 = 150 kvar
Q2 = 200·tan(acos 0.98) = 200·0.2031 = 40.61 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 109.4 kvar
Capacitance per phase
In a Δ bank each capacitor is connected between two lines.