Answer:4. Back emf fall but line current increases
Explanation:
Significance of Back EMF & Speed:
In the DC motor also generator action (mechanical energy converted into electrical energy) takes place. Due to generator action, the rotating conductor cuts the flux and emf induced, this induced emf is called back EMF.
It always opposes the supply voltage.
The induced back emf is A.C in the armature winding. But with respect to brushes, the back emf is D.C.
In a DC shunt motor, Back EMF is given by
Eb = V – IaRa
Where Ia is armature current
Ra is armature resistance
Also,
The back emf of a dc motor is directly proportional to speed.
Eb ∝ Nϕ
If the speed of a DC motor increases, there will be an increase in back emf also.
Current drawn in the DC motor is given by,
Ia = (V − Eb)/Ra
When the speed of a DC motor increases, emf will get an increase, and hence the current drawn will get decrease.
4. The direction of induced EMF in the DC motor is determined By the _____
Fleming’s Right-Hand Rule
Fleming’s Left-hand Rule
Maxwell Thumb Rule
Any of the above
Answer:1. Fleming’s Right-Hand Rule
Explanation:
The Fleming right-hand rule determines the direction of the induced EMF. According to Fleming’s Right-Hand Rule, if we hold our thumb, middle finger, and index finger of the right hand by an angle of 90°, then the index finger represents the direction of the magnetic field.
When the current-carrying conductor is placed in a magnetic field, the torque induces on the conductor, the torque rotates the conductor which cuts the flux of the magnetic field.
According to the Electromagnetic Induction Phenomenon “when the conductor cuts the magnetic field, EMF induces in the conductor”.
It is seen that the direction(Right-hand rule) of the induced emf is opposite to the applied voltage. Thereby the emf is known as the counter emf or back emf.
5. With the increase in speed of a D.C. motor back emf _______ but line current _______
Increase, Increase
Decrease, Increase
Decrease, Decrease
Increase, Decrease
Answer:4. Increase, Decrease
Explanation:
In the DC motor also generator action (mechanical energy converted into electrical energy) takes place. Due to generator action, the rotating conductor cuts the flux and emf induced, this induced emf is called back EMF.
It always opposes the supply voltage.
The induced back emf is A.C in the armature winding. But with respect to brushes, the back emf is D.C.
In a DC shunt motor, Back EMF is given by
Eb = V – IaRa
Where Ia is armature current
Ra is armature resistance
Also,
The back emf of a dc motor is directly proportional to speed.
Eb ∝ Nϕ
If the speed of a DC motor increases, there will be an increase in back emf also.
Current drawn in the DC motor is given by,
Ia = (V − Eb)/Ra
When the speed of a DC motor increases, emf will get an increase, and hence the current drawn will get decrease.
6. A 230 V dc motor has an armature circuit resistance of 0.6 Ω. If the full load armature current is 30 A and the no-load armature current is 4 A, the change in back emf from no load to full load will be
32.2 V
64 V
1.5 V
15.6 V
Answer:4. 15.6
Explanation:
In a DC shunt motor, Back EMF is given by
Eb = V – IaRa
Where, Ia is armature current
Ra is armature resistance
Application:
Given,
V = 230 volts
Ra = 0.6 Ω
Case 1: When the motor is under no load,
Ia = 4 A
∴ EbNL = 230 – (4 × 0.6) = 227.6 volts
Case 2: When the motor is under full load,
Ia = 30 A
∴ EbFL = 230 – (30 × 0.6) = 212 volts
Hence, change in back emf (ΔEb) from no load to full load will be,
ΔEb = 227.6 – 212 = 15.6 volts
7. A separately excited DC motor runs at 1000 rpm on no-load when its armature terminals are connected to a 200 V DC source and the rated voltage is applied to the field winding. The armature resistance of this motor is 1 Ω. The no-load armature current is negligible. With the motor developing its full load torque, the armature voltage is set so that the rotor speed is 500 rpm. When the load torque is reduced to 50% of the full load value under the same armature voltage conditions, the speed rises to 520 rpm. Neglecting the rational losses, the full load armature current (in Ampere) is _________.
4 A
8 A
2 A
16 A
Answer:2. 8 A
Explanation:
N1 = 1000 rpm
V = 200 V
Ra = 1 Ω
No load Back emf (Eb1) = 200 V
N2 = 500 rpm
T3 = 0.5 T2
N3 = 520 rpm
Eb1/Eb2 = 1000/500
200/Eb2 = 1000/500
Eb2 = 100 V
Eb1/Eb2 = 1000/500
⇒ Eb3 = 104
Eb3 = V – Ia3
104 = V – Ia3
T2/T3 ∝ Ia2/Ia3
T2/0.5T2 = Ia2/Ia3
Ia3 = Ia2/2 = 8A
8. A 10 hp, 240 V dc shunt motor, having armature-circuit resistance of 0.5 Ω and a full-load current of 40 A, is started by a starter, such that sections of required resistances in series with the armature-circuit should limit the starting current to 150% of the full-load current. The steady-state emf developed by the machine at full-load when the arm of the starter is moved to the next step is
120 V
100 V
80 V
60 V
Answer:3. 80 V
Explanation:
Given that, V = 240 V
Full load current I = 40 A
150% of full load current = 40 × 1.5 = 60 A
Starting resistance to limit current up to 150%
= 240/60 = 4 Ω
Back emf developed at full load when the starter is moved to next step = 240 – 40 × 4 = 80 V
9. The back EMF of DC motor is _______ of the applied voltage.
Equal
Half
Twice
Zero
Answer:2. Half
Explanation:
The back emf of the DC motor is equal to half of the applied voltage for maximum gross mechanical power. It is impossible to achieve this condition practically. Since half of the power input is wasted as heat in the armature, efficiency will be less than 50 percent taking account of other mechanical losses also.
10. If the field winding of a DC shunt motor gets disconnected while in normal operation then _______
Flux is reduced to Zero
Motor will run at high Speed
Motor gets damaged
All of the above
Answer:4. All of the above
Explanation:
If the field winding of a DC shunt motor gets disconnected while in normal operation then the flux will drop to almost zero value (residual value) and, therefore, the speed will increase to a tremendously high value and the motor may get damaged due to heavy centrifugal forces.
