PWM (Pulse Width Modulation) is often used for DC motor control using a microcontroller is Pulse Width Modulation (PWM) method. The speed of the electric motor depends on the modulator voltage. The greater the voltage, the faster the rotation of an electric motor.
The method is to apply a pulse train to the power terminals of the motor. The average voltage obtained at the terminals is then proportional to the duty cycle of the pulse train, which is proportional to the speed of rotation (rpm) of the motor.
Thus, as the duty cycle is increased, the motor rpm increases and vice versa. When the power supply is constant, it runs at 100% of its power rating (at no load).
As the duty cycle reduces, the speed and the power reduce. In the PWM technique duty ratio is a linear function with respect to speed.
2. The speed of DC shunt motor can be increased above its rated speed by
Increasing the supply voltage within its rated value
Increasing the flux per pole
Decreasing the flux per pole
Connecting additional resistance in series of the armature
Answer:3. Decreasing the flux per pole
Explanation:
Field control method:
By adding external resistance in the field circuit, we can control the speed of the DC shunt motor above the normal speed.
In this method, speed variation is accomplished by means of a variable resistance inserted in series with the shunt field.
An increase in controlling resistances reduces the field current with a reduction in flux and an increase in speed.
This method of speed control is independent of the load on the motor. Power wasted in controlling resistance is very less as field current is a small value.
We know that,
N ∝ Eb/φ
N ∝ (V − Ia R/φ
ϕ ↓ → N ↑
By varying flux, we can increase the speed more than its base speed. This method is a constant power and variable torque drive.
3. A separately excited dc motor has an armature resistance Ra = 0.05 Ω. The field excitation is kept constant. At an armature voltage of 100 V, the motor produces a torque of 500 Nm at zero speed. Neglecting all mechanical losses, the no-load speed of the motor (in radian/s) for an armature voltage of 150 V is
600 RPM
1200 RPM
300 RPM
900 RPM
Answer: 1. 600 RPM
Explanation:
At zero speed (N = 0)
induced voltage (E) = 0
T = KφIa1
⇒ T = KIa1 {∴ ϕ = Constant}
E = Vt − Ia1R1
Ia1 = 100 − 0/0.05 = 2000A
∴ K = T/Ia1 = 500/2000 = 1/4
At no -load condition, load torque (TL) = 0
E2Ia2 = Tω
⇒ Ia2 = 0
So, induced voltage
E2 = Kφω2 = Kω2 = 150
ω2 = E2/K = 150 × 4 /1
ω2 = 600 RPM
4. A 200 V d.c. shunt motor with armature resistance of 0.2 Ω and carrying a current of 50 A is running at 960 r.p.m. If the flux is reduced by 10% at constant torque and with negligible iron and friction losses, the speed will be near.
5. A 250 V dc shunt motor has an armature resistance of 0.2 Ω and a field resistance of 100 Ω. On no-load at rated voltage, it draws an armature current of 5 A and runs at 1200 RPM. When the motor is operating on load is it draws a total line current of 50 A at rated voltage, with a 5 % reduction in the air-gap flux due to armature reaction. The voltage drop across the brushes is 1 V. The speed of the motor, in rpm, under this loaded condition, is closest to:
6. A dc series motor with resistance between terminals of 1 Ω, runs at 800 RPM from a 200 V supply taking 15 A. If the speed is to be reduced to 475 rpm for the same supply voltage and current the additional series resistance to be inserted would be approximately
Therefore, Extra Resistance to be added in Series:
Rex = R2 – R1
= 6 – 1 = 5 Ω
7. The DC Motor, which can provide zero speed regulation at full load without any controller is
Series motor
Shunt motor
Cumulative compound
Differential compound
Answer: 2. Shunt motor
Explanation:
DC shunt motor has almost constant speed characteristics from no load to full load hence it provides zero speed regulation at full load without any controller.
8. Consider the following statements related to the speed control of a D.C. motor
(a) Speed may be controlled by changing the pole flux.
(b) Speed may be changed by changing voltage across the armature
Only (a) is correct
Only (b) is correct
Both (a) and (b) are correct
Both (a) and (b) are not correct
Answer:3. Both (a) and (b) are correct
Explanation:
Armature resistance control:
In a shunt motor, flux is constant when applied terminal voltage and shunt field resistance are constant. Therefore, speed of the motor is directly proportional to induced emf.
N ∝ Eb/φ
Eb = V – Ia Ra
The value of Eb depends upon the drop in the armature circuit.
We can control the speed by increasing armature resistance.
This method gives only below base speeds.
This method is a constant torque and variable power drive.
Field control method:
By adding external resistance in the field circuit, we can control the speed of the DC shunt motor above the normal speed.
In this method, speed variation is accomplished by means of a variable resistance inserted in series with the shunt field.
An increase in controlling resistances reduces the field current with a reduction in flux and an increase in speed.
This method of speed control is independent of the load on the motor. Power wasted in controlling resistance is very less as field current is a small value.
We know that,
N ∝ Eb/φ
N ∝ (V − Ia R/φ
ϕ ↓ → N ↑
By varying flux, we can increase the speed more than its base speed. This method is a constant power and variable torque drive.
9. A 240 V shunt motor with the armature resistance of 0.1 Ω runs at 850 r.p.m. for an armature current of 70 A. If its speed is to be reduced to 650 r.p.m., the resistance to be placed in series for an armature current of 50 A is nearly
0.82 Ω
1.13 Ω
1.24 Ω
1.34 Ω
Answer: 2. 1.13Ω
Explanation:
Given V = 240 V
Ra = 0.1 Ω
Armature current Ia = 70 A
Eb = 240 – 70 × 0.1 = 233 volt
Knφ = 233/850
Now when it runs on 650 rpm
Back emf Eb = (233 × 650)/850
Current flows through armature, Ia = 50 A
R = (240 − 178.17)/50 = 1.23 Ohm
External resistance, Rext = R – Ra = 1.23 – 0.1 = 1.13 Ω
10. A 250 V dc shunt machine has armature circuit resistance of 0.6 Ω and field circuit resistance of 125 Ω. The machine is connected to 250 V supply mains. The motor is operated as a generator and then as a motor separately. The line current of the machine in both cases is 50 A. The ratio of the speed as a generator to the speed as a motor is _______
1.27
1.30
1.20
1.17
Answer:1. 1.27
Explanation:
Terminal voltage VL = 250 V
Armature resistance Ra = 0.6 Ω
Field resistance Rf = 125 Ω
Line current in both IL = 50 A
Machine is running as a generator:
Field current, Ish = VL/Rf
= 250/125 = 2A
Armature current Ia = IL + Ish
⇒ Ia = 52 A
Induced emf: Eg = VL + Ia Ra
= 250 + 52 × 0.6
Eg = 281.2 V
Machine is running as motor:
Field current, Ish = VL/Rf
= 250/125 = 2A
Armature current Ia = IL – Ish
= 50 – 2 = 48 A
Back emf:
Eb = VL – IaRa = 250 – 48 × 0.6
Eb = 221.2 V
For DC shunt machine, this is constant hence
E ∝ N
= Eg/Eb = Ng/Nm
Ng/Nm = 281.2/221.2 = 1.27
11. A shunt DC motor’s Armature resistance is 1 ohm and field coil resistance is 100 ohms. The back emf constant is 9 volts for 100 RPM. The DC supply voltage is 100 V and it is running at a speed of 1000 RPM. What is DC supply current?
11 A
10 A
1 A
0.1 A
Answer: 1. 11 A
Explanation:
Armature resistance, Ra = 1 Ω
Field resistance, Rsh = 100Ω
Speed of motor, N = 1000 rpm
Supply voltage, Vs = 100 V
Back emf constant, K = 9 V for 100 rpm
= 0.09 V / rpm
∴ Back emf (Eb) can be calculated as
Eb = kN
∴ Eb = 0.09 × 1000
∴ Eb = 90 V
Now shunt field current can be calculated as
Ish = Vs/Rsh = 100/100
∴ Ish = 1 A
Now armature current can be calculated as
Vs = Eb + Ia Ra
Ia = (Vs − Eb)/Ra = (100 − 90)/1
Ia = 10 A
Supply current (Is) can be calculated as
I = Ia + Ish = 10 + 1
I = 11 A
12. A separately excited 300 V DC shunt motor under no-load runs at 900 rpm drawing an armature current of 2 A. The armature resistance is 0.5 Ω and leakage inductance is 0.01 H. When loaded, the armature current is 15 A. Then the speed in rpm is
840 RPM
880 RPM
850 RPM
830 RPM
Answer: 2. 880 RPM
Explanation:
Given
Ra = 0.5 Ω
Ia = 2A, leakage inductance, L = 0.01 H
V = 300 V
Eno load = V – IaRa
= 300 – 2 (0.5)
= 300 – 1 = 299 V
When load is applied, we have
Ra = 0.5 Ω & Ia = 15 A
The circuit can be drawn is shown in fig (b)
Hence, Eload – V – IaRa
= 300 – 15 × 0.5 = 292.5 V
E ∝ N
En0-load/Eload = Nn0-load/Nload
= 299/292.5 = 900/Nload
⇒ Nload = 880.434 rpm
13. A 120 V DC shunt motor takes 2 A at no load. It takes 7 A on full load while running at 1200 rpm. The armature resistance is 0.8 Ω and the shunt field resistance is 240 Ω. The no-load speed, in rpm is
2141 RPM
1241 RPM
1125 RPM
1535 RPM
Answer: 2. 1241 RPM
Explanation:
Voltage (V) = 120 V
shunt field resistance (Rsh) = 240 Ω
Shunt field current (Ish) = 120/240 = 0.5 A
Armature resistance (Ra) = 0.8 Ω
No load:
Load current (ILO) = 2 A
Armature current (Iao) = ILo – Ish
= 2 – 0.5 = 1.5 A
Let No load speed is No
Back emf (Ebo) = V – Iao Ra = 120 – (1.5) (0.8) = 118.8 V
14. A dc series motor of resistance 1 Ω across terminals runs at 1000 rpm at 250 V taking a current of 20 A. When an additional resistance of 6 Ω is inserted in series and taking the same current, the new speed would be
15. A 230-V DC shunt motor has an armature resistance of 0.25 Ω and runs at 1100 rpm, taking an armature current 40 A. It is desired to reduce the speed to 750 rpm. If the armature current remains the same, find the additional resistance to be connected in series with the armature circuit.
