34A shunt motor has a rated armature current of 40A whe The rated speed of the motor is 1000 rpm. The armature resis the speed of the motor if total torque is reduced to 70% of tha
34A shunt motor has a rated armature current of 40A whe The rated speed of the motor is 1000 rpm. The armature resis the speed of the motor if total torque is reduced to 70% of tha
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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FUFULAR PUBLICATIUNS
Rated filll I
The circuit diagram is dra
At starting, the speed of the motor is zero, hence back emf developed is zero
(E,
P¢Z,
where, n is the speed of the motor).
The field current, I, = 120
R, 60 A=2A
In the absence of E,, the armature current is very high, which is potentially detriments
to the armature circuit. A starter is an additional resistance which limits the startim
current and ensures smooth operation of the motor, hence it is necessary.
1,-1, +1,
1=40 -2- 38 A
aack emf E, =V, -v.- Brush contact d
34A shunt motor has a rated armature current o
VThe rated speed of the motor is 1000 rpm. The armature resistance is 0.2 ohm, Find
the speed of the motor if total torque is reduced to 70% of that at rated load and
ohm resistance is inserted in series with the armature.
me flux
connected to 200y
e of zero
al cmf or
-(120-02x 38-3)V=(120-76-3)V =109 4V
from the
[WBUT 2013]
POZN
E.60A
Answer:
At rated load
PO E 109.4
Virpm
60A N 1800
onductors.
e armature
ne full load
WBUT 2012]
V= 200
E, =V, -1r, =(200 - 0.2x10) V- (200 - 2) V =198 V
Torque is proportional to current.
Let 1, be the rated current at full load,
Let 1, be the current when torque is 70% of that at rated load.
or.
At half load, line current =A= 20 A.
The ficld current remains constant for constant terminal voltage ie, 2A,
Hence, I = (20 - 2) A=18A
The new back emf
E -V, - I'r, - Brush contact drop
T 1,
1, = 1,-10x0.7 = 7A
=(120 - 0.2x18-3)V= (120-3.6-3)V=113AV
or,
E PO 1094
N' 604 1800
(The flux remains constant for constant field current)
1800 113.4x1800
Hence back e.m.f = {200-(3+ 0.2)× 7} V =177.6V
(Since back emfis proportional to speed at constant flux]
N,
N' = E
109 4 pm = 1865.8 rpm=1865 rpan
N = N, x 1000 x177.6
E
109.4
rpm = 897 rpm
125
At 125% of full load, load current = 40XA 50A
100
198
3.5. a) A 120 V D.C. Shunt Motor having an armature resistance of 0.2 n and field
resistance of 60 0, draws a line current of 40 A at full load, The brush voltage drop
is 3 V and the rated full load speed is 1800 rpm. Calculate the speed at half load
and 125% of full load.
(WBUT 2012)
The armature current, I = 50-2= 48 A
The back emf= (120– 48x0.2-3)V=(120-9.6 –3)V
E =107.4 V
E Po_109.4
N' 604 1800
The relation between
, where V, is the
e current and , the
[WBUT 2014]
Answer:
The relevant data for the D.C. shunt motor are listed below:
Terminal voltage
V =120 V
I- 40A
Armature resistance
fihe flux remains constant for constant field current and terminal voltage)
R, = 0.20
R, = 602
107.4x1800
1094 rpm = 1767 rpm
R-20
Field resistance
V = 120V
Ry = 602
Line current
= 40 A
Brush voltage drop 3 V
BEE-143](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F50db262f-31a4-47fe-992a-58e04ef6be2d%2Fb9deafb0-997a-4e90-8708-0b5640d49e49%2F7seokuv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Scanned by Car
trickstarvivek.com
NG
FUFULAR PUBLICATIUNS
Rated filll I
The circuit diagram is dra
At starting, the speed of the motor is zero, hence back emf developed is zero
(E,
P¢Z,
where, n is the speed of the motor).
The field current, I, = 120
R, 60 A=2A
In the absence of E,, the armature current is very high, which is potentially detriments
to the armature circuit. A starter is an additional resistance which limits the startim
current and ensures smooth operation of the motor, hence it is necessary.
1,-1, +1,
1=40 -2- 38 A
aack emf E, =V, -v.- Brush contact d
34A shunt motor has a rated armature current o
VThe rated speed of the motor is 1000 rpm. The armature resistance is 0.2 ohm, Find
the speed of the motor if total torque is reduced to 70% of that at rated load and
ohm resistance is inserted in series with the armature.
me flux
connected to 200y
e of zero
al cmf or
-(120-02x 38-3)V=(120-76-3)V =109 4V
from the
[WBUT 2013]
POZN
E.60A
Answer:
At rated load
PO E 109.4
Virpm
60A N 1800
onductors.
e armature
ne full load
WBUT 2012]
V= 200
E, =V, -1r, =(200 - 0.2x10) V- (200 - 2) V =198 V
Torque is proportional to current.
Let 1, be the rated current at full load,
Let 1, be the current when torque is 70% of that at rated load.
or.
At half load, line current =A= 20 A.
The ficld current remains constant for constant terminal voltage ie, 2A,
Hence, I = (20 - 2) A=18A
The new back emf
E -V, - I'r, - Brush contact drop
T 1,
1, = 1,-10x0.7 = 7A
=(120 - 0.2x18-3)V= (120-3.6-3)V=113AV
or,
E PO 1094
N' 604 1800
(The flux remains constant for constant field current)
1800 113.4x1800
Hence back e.m.f = {200-(3+ 0.2)× 7} V =177.6V
(Since back emfis proportional to speed at constant flux]
N,
N' = E
109 4 pm = 1865.8 rpm=1865 rpan
N = N, x 1000 x177.6
E
109.4
rpm = 897 rpm
125
At 125% of full load, load current = 40XA 50A
100
198
3.5. a) A 120 V D.C. Shunt Motor having an armature resistance of 0.2 n and field
resistance of 60 0, draws a line current of 40 A at full load, The brush voltage drop
is 3 V and the rated full load speed is 1800 rpm. Calculate the speed at half load
and 125% of full load.
(WBUT 2012)
The armature current, I = 50-2= 48 A
The back emf= (120– 48x0.2-3)V=(120-9.6 –3)V
E =107.4 V
E Po_109.4
N' 604 1800
The relation between
, where V, is the
e current and , the
[WBUT 2014]
Answer:
The relevant data for the D.C. shunt motor are listed below:
Terminal voltage
V =120 V
I- 40A
Armature resistance
fihe flux remains constant for constant field current and terminal voltage)
R, = 0.20
R, = 602
107.4x1800
1094 rpm = 1767 rpm
R-20
Field resistance
V = 120V
Ry = 602
Line current
= 40 A
Brush voltage drop 3 V
BEE-143
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