A buck converter turns 48 V into 24 V for a 25 Ω load at 200 kHz, with L = 330 µH and C = 10 µF. Find the peak-to-peak output voltage ripple ΔVo in mV.
A DC motor delivers 5.5 kW of mechanical power at 3000 rpm. Find the shaft torque T.
P = 5.5 kWn = 3000 rpm
T = N·m
4.DC motorsHard
A 120 V DC motor has an armature resistance of 0.5 Ω. Find the series starting resistance Rst that limits the starting current to 80 A.
V = 120 VRa = 0.5 ΩImax = 80 A
Rst = Ω
5.AC power & power factorMedium
Two loads are connected in parallel to a 230 V single-phase supply: a 10 kW load at a power factor of 0.75 lagging and a 3 kW heater (unity power factor). Find the total current I drawn from the supply.
A 12 V (rms), 60 Hz source has an internal impedance Zs = 25 + j40 Ω. The load impedance ZL can be chosen freely. Find the maximum real power Pmax that can be delivered to the load.
Vs = 12 V (rms)f = 60 HzZs = 25 + j40 Ω
Pmax = W
7.Three-phase powerMedium
A balanced Δ-connected load with an impedance of 16 + j12 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 = 16 + j12 Ω per phase
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 80 kW at a power factor of 0.65 lagging from a 400 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.95 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.
P = 80 kWpf1 = 0.65pf2 = 0.95VL = 400 Vf = 60 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: Imax = 3.706 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 = 25.2/36 = 0.7
Average and ripple current
Io = 25.2/8 = 3.15 A
ΔIL = VL·D/(L·f) = 10.8·0.7/(68·10⁻⁶·100·10³) = 1.112 A
Peak current
Imax = 3.15 + 1.112/2 = 3.706 A
Answer
Imax = 3.706 A
2.DC-DC converters
Answer: ΔVo = 11.36 mV
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 = 24/48 = 0.5
Inductor ripple current
ΔIL = (48 − 24)·0.5/(330·10⁻⁶·200·10³) = 0.1818 A
Output ripple
ΔVo = 0.1818/(8·200·10³·10·10⁻⁶) = 11.36 mV
Answer
ΔVo = 11.36 mV
3.DC motors
Answer: T = 17.51 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·3000/60 = 314.2 rad/s
Torque
T = P/ω = 5500/314.2 = 17.51 N·m
Answer
T = 17.51 N·m
4.DC motors
Answer: Rst = 1 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.
Total resistance
Ra + Rst = 120/80 = 1.5 Ω
Starting resistor
Rst = 1.5 − 0.5 = 1 Ω
Answer
Rst = 1 Ω
The resistor is cut out step by step as the motor speeds up and the back-EMF grows.
5.AC power & power factor
Answer: I = 68.3 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 of each load
A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).
Q1 = 10·tan(arccos 0.75) = 8.819 kvar
Q2 = 0 = 0 kvar
Totals
P = 10 + 3 = 13 kW
Q = 8.819 + 0 = 8.819 kvar
S = √(13² + 8.819²) = 15.71 kVA
Total current
I = S/V = 15709/230 = 68.3 A
Answer
I = 68.3 A
6.AC power & power factor
Answer: Pmax = 1.44 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* = 25 − j40 Ω
Maximum power
I = Vs/(2Rs) = 12/50 = 0.24 A
Pmax = I²·RL = Vs²/(4Rs) = 12²/(4·25) = 1.44 W
Answer
Pmax = 1.44 W
7.Three-phase power
Answer: P = 20.67 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 = VL = 415 V
Phase current
|Z| = √(16² + 12²) = 20 Ω
Iph = 415/20 = 20.75 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·20.75²·16 = 20.67 kW
Answer
P = 20.67 kW
Check: cos φ = R/|Z| = 0.8, and √3·VL·IL·cos φ gives the same power.
8.Three-phase power
Answer: Cph = 371.6 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 = 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
Capacitance per phase
In a Δ bank each capacitor is connected between two lines.