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EM-Problems
First Semester 2019-2020
Problem 1
An 8-pole, wave-connected armature has 600 conductors and is driven at 625rev/min. If the flux
per pole is 20mWb.
- Determine the generated e.m.f.
Solution:
๐’ = ๐Ÿ”๐ŸŽ๐ŸŽ, ๐’„ = ๐Ÿ (๐’‡๐’๐’“ ๐’‚ ๐’˜๐’‚๐’—๐’† ๐’˜๐’Š๐’๐’…๐’Š๐’๐’ˆ), ๐’‘ = ๐Ÿ’ ๐’‘๐’‚๐’Š๐’“๐’”
๐’=
๐Ÿ”๐Ÿ๐Ÿ“ ๐’“๐’†๐’—
[
],
๐Ÿ”๐ŸŽ
๐’”
๐“ = ๐Ÿ๐ŸŽ ∗ ๐Ÿ๐ŸŽ−๐Ÿ‘ ๐‘พ๐’ƒ
๐‘ฎ๐’†๐’๐’†๐’“๐’‚๐’•๐’†๐’… ๐’†. ๐’Ž. ๐’‡. , ๐‘ฌ =
๐Ÿ๐‘ท๐“๐’๐’
๐’„
๐Ÿ”๐Ÿ๐Ÿ“
๐Ÿ (๐Ÿ’) (๐Ÿ๐ŸŽ ∗ ๐Ÿ๐ŸŽ−๐Ÿ‘ ) (
) (๐Ÿ”๐ŸŽ๐ŸŽ)
๐Ÿ”๐ŸŽ
๐‘ฌ=
= ๐Ÿ“๐ŸŽ๐ŸŽ ๐‘ฝ๐’๐’๐’•๐’”
๐Ÿ
Problem 2
A 4-pole generator has a lap-wound armature with 50 slots with 16 conductors per slot.
The useful flux per pole is 30mWb.
- Determine the speed at which the machine must be driven to generate an e.m.f. of 240V.
Solution:
๐ธ = 240 ๐‘‰, ๐‘ = 2๐‘ ( ๐‘“๐‘œ๐‘Ÿ ๐‘Ž ๐‘™๐‘Ž๐‘ ๐‘ค๐‘–๐‘›๐‘‘๐‘–๐‘›๐‘”), ๐‘ = 50 ∗ 16 = 800, ๐œ™ = 30 ∗ 10−3 ๐‘Š๐‘
๐‘ฎ๐’†๐’๐’†๐’“๐’‚๐’•๐’†๐’… ๐’†. ๐’Ž. ๐’‡. , ๐‘ฌ =
๐‘น๐’†๐’‚๐’“๐’“๐’‚๐’๐’ˆ๐’Š๐’๐’ˆ ๐’ˆ๐’Š๐’—๐’†๐’”, ๐’”๐’‘๐’†๐’†๐’…, ๐’ =
๐Ÿ๐’‘๐“๐’๐’
๐Ÿ๐’‘๐“๐’๐’
=
= ๐“๐’๐’
๐’„
๐Ÿ๐’‘
๐‘ฌ
๐Ÿ๐Ÿ’๐ŸŽ
๐’“๐’†๐’—
๐’“๐’†๐’—
=
=
๐Ÿ๐ŸŽ
๐’๐’“
๐Ÿ”๐ŸŽ๐ŸŽ
(๐Ÿ‘๐ŸŽ ∗ ๐Ÿ๐ŸŽ−๐Ÿ‘ ) (๐Ÿ–๐ŸŽ๐ŸŽ)
๐“๐’
๐’”
๐’Ž๐’Š๐’
Problem 3
An 8-pole, lap-wound armature has 1200 conductors and a flux per pole of 0.03Wb.
- Determine the e.m.f. generated when running at 500rev/min.
Solution:
๐‘ฎ๐’†๐’๐’†๐’“๐’‚๐’•๐’†๐’… ๐’†. ๐’Ž. ๐’‡. , ๐‘ฌ =
๐Ÿ๐‘ท๐“๐’๐’
๐Ÿ๐‘ท๐“๐’๐’
=
= ๐“๐’๐’
๐’„
๐Ÿ๐‘ท
๐‘ช = ๐’‘ ๐’‡๐’๐’“ ๐’๐’‚๐’‘ − ๐’˜๐’๐’–๐’๐’… ๐’Ž๐’‚๐’„๐’‰๐’Š๐’๐’†
๐‘ฌ = ๐“ ๐’ ๐’ = (๐ŸŽ. ๐ŸŽ๐Ÿ‘) (
๐Ÿ“๐ŸŽ๐ŸŽ
) (๐Ÿ๐Ÿ๐ŸŽ๐ŸŽ) = ๐Ÿ‘๐ŸŽ๐ŸŽ ๐’—๐’๐’๐’•๐’”
๐Ÿ”๐ŸŽ
1
Problem 4
- Determine the generated e.m.f. in Problem 3 if the armature is wave-wound.
Solution:
๐บ๐‘’๐‘›๐‘’๐‘Ÿ๐‘Ž๐‘ก๐‘’๐‘‘ ๐‘’. ๐‘š. ๐‘“. , ๐ธ =
2๐‘ƒ๐œ™๐‘›๐‘
2๐‘ƒ๐œ™๐‘›๐‘
=
๐‘
2
"๐’„ = ๐Ÿ ๐’‡๐’๐’“ ๐’˜๐’‚๐’—๐’† − ๐’˜๐’๐’–๐’๐’… ๐’Ž๐’‚๐’„๐’‰๐’Š๐’๐’†"
๐‘ฌ = ๐‘ท ๐“ ๐’ ๐’ = (๐Ÿ’) (๐“ ๐’ ๐’) = ๐Ÿ‘๐ŸŽ๐ŸŽ ๐’—๐’๐’๐’•๐’”
๐‘ฌ = (๐Ÿ’) (๐Ÿ‘๐ŸŽ๐ŸŽ) = ๐Ÿ๐Ÿ๐ŸŽ๐ŸŽ ๐’—๐’๐’๐’•๐’”
Problem 5
A d.c. shunt-wound generator running at constant speed generates a voltage of 150 V at a
certain value of field current.
- Determine the change in the generated voltage when the field current is reduced by 20%,
assuming the flux is proportional to the field current.
Solution:
The generated e.m.f. E of a generator is proportional to ะค ω, i.e. is proportional to ะค n,
where ะค is the flux and n is the speed of rotation. It follows that E = k ะค n, where k is a
constant.
At speed n1 and flux ะค1, E1 = k ะค1 n1
At speed n2 and flux ะค2, E2 = k ะค2 n2
Thus, by division:
๐ธ1
๐‘˜ ๐œ™1 ๐‘›1
๐œ™1 ๐‘›1
=
=
๐ธ2
๐‘˜ ๐œ™2 ๐‘›2
๐œ™2 ๐‘›2
The initial conditions are E1 =150V, ะค = ะค1 and n = n1. When the flux is reduced by 20%, the
new value of flux is 80/100 or 0.8 of the initial value, i.e. ะค2 = 0.8 ะค1.
Since the generator is running at constant speed, n2 = n1.
๐‘‡โ„Ž๐‘ข๐‘ 
๐‘กโ„Ž๐‘Ž๐‘ก ๐‘–๐‘ ,
๐ธ1
๐œ™1 ๐‘›1
๐œ™1 ๐‘›1
1
=
=
=
๐ธ2
๐œ™2 ๐‘›2
0.8 ๐œ™1 ๐‘›1
0.8
๐ธ2 = 150 ∗ 0.8 = 120 ๐‘‰
Thus, a reduction of 20% in the value of the flux reduces the generated voltage to 120 V at
constant speed.
2
Problem 6
A d.c. generator running at 30 rev/s generates an e.m.f. of 200V.
- Determine the percentage increase in the flux per pole required to generate 250 V at 20 rev/s.
Solution:
E ∝ ะค ω and since ω =2πn, E ∝ n.
Let E1 = 200 V, n1 = 30 rev/s and flux per pole at this speed be ะค1
Let E2 = 250 V, n2 = 20 rev/s and flux per pole at this speed be ะค2
๐‘†๐‘–๐‘›๐‘๐‘’ ๐ธ ∝ ๐œ™ ๐‘› ๐‘กโ„Ž๐‘’๐‘›
๐ป๐‘’๐‘›๐‘๐‘’
๐‘“๐‘Ÿ๐‘œ๐‘š ๐‘คโ„Ž๐‘–๐‘โ„Ž, ๐œ™2 =
๐ธ1
๐œ™1 ๐‘›1
=
๐ธ2
๐œ™2 ๐‘›2
200
๐œ™1 (30)
=
250
๐œ™2 (20)
๐œ™1 (30) (250)
= ๐Ÿ. ๐Ÿ–๐Ÿ•๐Ÿ“ ๐“๐Ÿ
(20) (200)
Hence the increase in flux per pole needs to be 87.5 %
Problem 7
- Determine the terminal voltage of a generator which develops an e.m.f. of 200 V and has an
armature current of 30 A on load. Assume the armature resistance is 0.30 โ„ฆ
Solution:
With reference to the Figure, terminal voltage,
VT = E – (Ia * Ra) = 200 – (30) (0.30) = 200 – 9 = 191 Volts
Problem 8
A generator is connected to a 60 โ„ฆ load and a current of 8 A flows. If the armature resistance
is 1โ„ฆ
- Determine (a) the terminal voltage, and (b) the generated e.m.f.
Solution:
a- Terminal voltage, V = Ia * RL = (8) (60) = 480 Volts
b- Generated e.m.f., E = V + (Ia* Ra) = 480 + (8 *1) = 480 + 8 = 488 Volts
3
Problem 9
A separately-excited generator develops a no-load e.m.f. of 150V at an armature speed of 20
rev/s and a flux per pole of 0.10Wb.
Determine the generated e.m.f. when: (a) the speed increases to 25 rev/s and the pole flux remains unchanged
(b) the speed remains at 20 rev/s and the pole flux is decreased to 0.08Wb.
(c) the speed increases to 24 rev/s and the pole flux is decreased to 0.07Wb.
Solution:
๐‘“๐‘Ÿ๐‘œ๐‘š ๐‘คโ„Ž๐‘–๐‘โ„Ž,
Hence,
๐‘“๐‘Ÿ๐‘œ๐‘š ๐‘คโ„Ž๐‘–๐‘โ„Ž,
๐ธ2 =
๐ธ1
∅1 ๐‘›1
=
๐ธ2
∅2 ๐‘›2
(0.10)(20)
150
=
(0.10)(25)
๐ธ2
(150)(0.10)(25)
= 187.5 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
(0.10)(20)
(0.10)(20)
150
=
(0.08)(20)
๐ธ3
๐‘“๐‘Ÿ๐‘œ๐‘š ๐‘คโ„Ž๐‘–๐‘โ„Ž, ๐‘’. ๐‘š. ๐‘“. ,
๐ธ3 =
(150)(0.08)(20)
= 120 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
(0.10)(20)
(0.10) (20)
150
=
(0.07) (24)
๐ธ4
๐‘“๐‘Ÿ๐‘œ๐‘š ๐‘คโ„Ž๐‘–๐‘โ„Ž, ๐‘’. ๐‘š. ๐‘“. ,
๐ธ4 =
4
(150)(0.07)(24)
= 126 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
(0.10)(20)
Problem 10
A shunt generator supplies a 20 kW load at 200 V through cables of resistance, R= 100 mโ„ฆ.
If the field winding resistance, Rf =50 โ„ฆ and the armature resistance, Ra =40 mโ„ฆ,
Determine: (a) the terminal voltage
(b) the e.m.f. generated in the armature.
Solution:
(a) The circuit is as shown in the Figure.
๐ฟ๐‘œ๐‘Ž๐‘‘ ๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก, ๐ผ =
20000 ๐‘Š๐‘Ž๐‘ก๐‘ก๐‘ 
= 100 ๐ด
200 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
๐‘‰๐‘œ๐‘™๐‘ก ๐‘‘๐‘Ÿ๐‘œ๐‘ ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘๐‘Ž๐‘๐‘™๐‘’๐‘  ๐‘ก๐‘œ ๐‘กโ„Ž๐‘’ ๐‘™๐‘œ๐‘Ž๐‘‘ = ๐ผ ∗ ๐‘… = (100 ๐ด) ∗ (100 ∗ 10−3 ๐›บ) = 10 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
๐ป๐‘’๐‘›๐‘๐‘’ ๐‘ก๐‘’๐‘Ÿ๐‘š๐‘–๐‘›๐‘Ž๐‘™ ๐‘ฃ๐‘œ๐‘™๐‘ก๐‘Ž๐‘”๐‘’, ๐‘‰ = 200 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  + 10 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  = 210 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
๐ด๐‘Ÿ๐‘š๐‘Ž๐‘ก๐‘ข๐‘Ÿ๐‘’ ๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก ๐ผ๐‘Ž = ๐ผ๐‘“ + ๐ผ
๐น๐‘–๐‘’๐‘™๐‘‘ ๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก, ๐ผ๐‘“ =
๐‘‰
210 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
=
= 4.2 ๐ด
๐‘…๐‘“
50 Ω
๐ป๐‘’๐‘›๐‘๐‘’ ๐ผ๐‘Ž = ๐ผ๐‘“ + ๐ผ = 4.2 ๐ด + 100 ๐ด = 104.2 ๐ด
๐บ๐‘’๐‘›๐‘’๐‘Ÿ๐‘Ž๐‘ก๐‘’๐‘‘ ๐‘’. ๐‘š. ๐‘“.
๐ธ = ๐‘‰ + ๐ผ๐‘Ž ๐‘…๐‘Ž
๐ธ = 210 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  + (104.2 ๐ด) (40 ∗ 10−3 Ω) = 210 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  + 4.168 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  = 214.17 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
5
Problem 11
A short-shunt compound generator supplies 80 A at 200 V. If the field resistance, Rf = 40 โ„ฆ, the
series resistance, Rse = 0.02 โ„ฆ and the armature resistance, Ra = 0.04 โ„ฆ, determine the e.m.f.
generated.
Solution:
The circuit is shown in the Figure.
๐‘‰๐‘œ๐‘™๐‘ก ๐‘‘๐‘Ÿ๐‘œ๐‘ ๐‘–๐‘› ๐‘ ๐‘’๐‘Ÿ๐‘–๐‘’๐‘  ๐‘ค๐‘–๐‘›๐‘‘๐‘–๐‘›๐‘” = ๐ผ ๐‘…๐‘†๐‘’ = (80 ๐ด) (0.02 Ω) = 1.6 ๐‘‰
๐‘ƒ๐‘œ๐‘ก๐‘’๐‘›๐‘ก๐‘–๐‘Ž๐‘™ ๐‘‘๐‘–๐‘“๐‘“๐‘’๐‘Ÿ๐‘’๐‘›๐‘๐‘’ ๐‘Ž๐‘๐‘Ÿ๐‘œ๐‘ ๐‘  ๐‘กโ„Ž๐‘’ ๐‘“๐‘–๐‘’๐‘™๐‘‘ ๐‘ค๐‘–๐‘›๐‘‘๐‘–๐‘›๐‘” = ๐‘๐‘œ๐‘ก๐‘’๐‘›๐‘ก๐‘–๐‘Ž๐‘™ ๐‘‘๐‘–๐‘“๐‘“๐‘’๐‘Ÿ๐‘’๐‘›๐‘๐‘’ ๐‘Ž๐‘๐‘Ÿ๐‘œ๐‘ ๐‘  ๐‘Ž๐‘Ÿ๐‘š๐‘Ž๐‘ก๐‘ข๐‘Ÿ๐‘’
๐‘‰1 = 200 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  + 1.6 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  = 201.6 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
๐น๐‘–๐‘’๐‘™๐‘‘ ๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก ๐ผ๐‘“ =
๐‘‰1
๐‘…๐‘“
=
201.6 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
= 5.04 ๐ด
40 Ω
๐ด๐‘Ÿ๐‘š๐‘Ž๐‘ก๐‘ข๐‘Ÿ๐‘’ ๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก, ๐ผ๐‘Ž = ๐ผ + ๐ผ๐‘“ = 80 ๐ด + 5.04 ๐ด = 85.04 ๐ด
๐บ๐‘’๐‘›๐‘’๐‘Ÿ๐‘Ž๐‘ก๐‘’๐‘‘ ๐‘’. ๐‘š. ๐‘“. , ๐ธ = ๐‘‰1 + (๐ผ๐‘Ž ∗ ๐‘…๐‘Ž )
๐ธ = 201.6 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  + (85.04 ๐ด ∗ 0.04 Ω) = 201.6 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  + 3.4 ๐‘ฃ๐‘œ๐‘™๐‘ก๐‘  = 205 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
Problem 12
A 10kW shunt generator having an armature circuit resistance of 0.75 โ„ฆ and a field resistance of
125 โ„ฆ, generates a terminal voltage of 250V at full load.
- Determine the efficiency of the generator at full load, assuming the iron, friction and windage
losses amount to 600W.
Solution:
The circuit is shown in the Figure.
๐‘‚๐‘ข๐‘ก๐‘๐‘ข๐‘ก ๐‘๐‘œ๐‘ค๐‘’๐‘Ÿ = 10000 ๐‘Š = ๐‘‰ ∗ ๐ผ
๐‘“๐‘Ÿ๐‘œ๐‘š ๐‘คโ„Ž๐‘–๐‘โ„Ž, ๐‘™๐‘œ๐‘Ž๐‘‘ ๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก
๐น๐‘–๐‘’๐‘™๐‘‘ ๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก, ๐ผ๐‘“ =
๐‘‰
๐‘…๐‘“
=
๐ผ=
250 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
125 Ω
10000
๐‘‰
=
10000 ๐‘Š๐‘Ž๐‘ก๐‘ก
250 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
= 40 ๐ด
=2๐ด
๐ด๐‘Ÿ๐‘š๐‘Ž๐‘ก๐‘ข๐‘Ÿ๐‘’ ๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก, ๐ผ๐‘Ž = ๐ผ๐‘“ + ๐ผ = 2 ๐ด + 40 ๐ด = 42 ๐ด
Efficiency,
๐‘‰∗๐ผ
๐œ‚= (
) ∗ 100 %
2
๐‘‰ ∗ ๐ผ + ๐ผ๐‘Ž ๐‘…๐‘Ž + ๐ผ๐‘“ ∗ ๐‘‰ + ๐ถ
๐œ‚= (
10000
10000
∗
100
%
=
∗ 100 % = 80.5 %
)
10000 + (422 )(0.75) + (2)(250) + 600
12423
6
Problem 13
A d.c. motor operates from a 240 V supply. The armature resistance is 0.2 โ„ฆ.
- Determine the back e.m.f. when the armature current is 50 A.
Solution:
๐น๐‘œ๐‘Ÿ ๐‘Ž ๐‘š๐‘œ๐‘ก๐‘œ๐‘Ÿ, ๐‘‰ = ๐ธ + ๐ผ๐‘Ž ∗ ๐‘…๐‘Ž
โ„Ž๐‘’๐‘›๐‘๐‘’ ๐‘๐‘Ž๐‘๐‘˜ ๐‘’. ๐‘š. ๐‘“. ,
๐ธ = ๐‘‰ − (๐ผ๐‘Ž ∗ ๐‘…๐‘Ž )
๐ธ = 240 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  − (50 ๐ด ∗ 0.2 Ω) = 240 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  − 10 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  = 230 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
Problem 14
The armature of a d.c. machine has a resistance of 0.25 โ„ฆ and is connected to a 300 V supply.
- Calculate the e.m.f. generated when it is running:
(a) as a generator giving 100 A.
(b) as a motor taking 80 A.
Solution:
(a) As a generator, generated e.m.f.,
๐ธ = ๐‘‰ + (๐ผ๐‘Ž ∗ ๐‘…๐‘Ž )
๐ธ = 300 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  + (100๐ด ∗ 0.25 Ω) = 300 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  + 25 ๐‘ฃ๐‘œ๐‘™๐‘ก๐‘  = 325 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
(b) As a motor, generated e.m.f. (or back e.m.f.),
๐ธ = ๐‘‰ − (๐ผ๐‘Ž ∗ ๐‘…๐‘Ž )
๐ธ = 300 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  − (80 ๐ด ∗ 0.25 Ω) = 280 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
Problem 15.
An 8-pole d.c. motor has a wave-wound armature with 900 conductors. The useful flux per
pole is 25 mWb. Determine the torque exerted when a current of 30A flows in each armature
conductor.
Solution:
P = 4, c = 2 for wave winding, ะค = 25 * 10-3 Wb, Z= 900, Ia = 30 A
(4)(25 ∗ 10−3 )(900)(30)
๐‘ ๐œ™ ๐‘ ๐ผ๐‘Ž
๐‘ก๐‘œ๐‘Ÿ๐‘ž๐‘ข๐‘’ ๐‘‡ =
=
= 429.7 ๐‘๐‘š
๐œ‹๐‘
๐œ‹(2)
7
Problem 16.
- Determine the torque developed by a 350V d.c. motor having an armature resistance of 0.5 โ„ฆ
and running at 15 rev/s. The armature current is 60 A.
Solution:
๐‘‰ = 350 ๐‘‰,
๐‘…๐‘Ž = 0.5 Ω,
๐‘› = 15
๐‘Ÿ๐‘’๐‘ฃ
,
๐‘ 
๐ผ๐‘Ž = 60 ๐ด
๐ต๐‘Ž๐‘๐‘˜ ๐‘’. ๐‘š. ๐‘“. ๐ธ = ๐‘‰ − (๐ผ๐‘Ž ∗ ๐‘…๐‘Ž ) = 350 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  − (60 ๐ด ∗ 0.5 Ω) = 320 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
๐‘ก๐‘œ๐‘Ÿ๐‘ž๐‘ข๐‘’ ๐‘‡ =
(320 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ ) (60 ๐ด)
๐ธ ๐ผ๐‘Ž
=
= 203.7 ๐‘๐‘š
๐‘Ÿ๐‘’๐‘ฃ
2๐œ‹๐‘›
2 ๐œ‹ (15 ๐‘  )
Problem 17
A six-pole lap-wound motor is connected to a 250V d.c. supply. The armature has 500 conductors
and a resistance of 1โ„ฆ. The flux per pole is 20mWb.
2p = one pair of poles
Calculate: -
This means
(a) the speed
C = number of poles
(b) the torque developed when the armature current is 40A.
Solution:
๐‘‰ = 250 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ , ๐‘ = 500, ๐‘…๐‘Ž = 1 Ω, Φ = 20 ∗ 10−3 Wb, ๐ผ๐‘Ž = 40 A, c = 2P for lap winding
๐ต๐‘Ž๐‘๐‘˜ ๐‘’. ๐‘š. ๐‘“. ๐ธ = ๐‘‰ − (๐ผ๐‘Ž ∗ ๐‘…๐‘Ž ) = 250 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  − (40 ๐ด ∗ 1 Ω) = 210 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
๐ธ. ๐‘š. ๐‘“.
2๐‘๐œ™๐‘›๐‘
2๐‘ (20 ∗ 10−3 ๐‘Š๐‘) ๐‘› (500)
๐ธ=
= ๐‘–. ๐‘’. 210 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  =
๐‘
2๐‘
๐ป๐‘’๐‘›๐‘๐‘’ ๐‘ ๐‘๐‘’๐‘’๐‘‘ ๐‘› =
210 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
๐‘Ÿ๐‘’๐‘ฃ
๐‘Ÿ๐‘’๐‘ฃ
(21
=
21
๐‘œ๐‘Ÿ
∗
60)
=
1260
(20 ∗ 10−3 ๐‘Š๐‘)(500)
๐‘ 
๐‘š๐‘–๐‘›
๐‘ก๐‘œ๐‘Ÿ๐‘ž๐‘ข๐‘’ ๐‘‡ =
(210 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ )(40 ๐ด)
๐ธ ๐ผ๐‘Ž
=
= 63.66 ๐‘๐‘š
๐‘Ÿ๐‘’๐‘ฃ
2๐œ‹๐‘›
2 ๐œ‹ (21 ๐‘  )
8
Problem 18
The shaft torque of a diesel motor driving a 100V d.c. shunt-wound generator is 25Nm. The
armature current of the generator is 16 A at this value of torque. If the shunt field regulator is
adjusted so that the flux is reduced by 15%, the torque increases to 35Nm.
- Determine the armature current at this new value of torque.
Solution:
The shaft torque T of a generator is proportional to Ia, where is the flux and Ia is the armature
current.
Thus, T = k ะค Ia, where k is a constant.
The torque at flux ะค1 and armature current Ia1 is T1 = k ะค1 Ia1.
Similarly, T2 = k ะค2 Ia2
๐‘‡1
๐‘˜ ∅1 ๐ผ๐‘Ž1
∅1 ๐ผ๐‘Ž1
=
=
๐‘‡2
๐‘˜ ∅2 ๐ผ๐‘Ž2
∅2 ๐ผ๐‘Ž2
๐ต๐‘ฆ ๐‘‘๐‘–๐‘ฃ๐‘–๐‘ ๐‘–๐‘œ๐‘›
๐ป๐‘’๐‘›๐‘๐‘’
๐‘–. ๐‘’.
25
∅1 ∗ 16
=
35
0.85 ∅1 ∗ ๐ผ๐‘Ž2
๐ผ๐‘Ž2 =
16 ∗ 35
= 26.35 ๐ด
0.85 ∗ 25
That is, the armature current at the new value of torque is 26.35 A.
9
Problem 19
A 100V d.c. generator supplies a current of 15A when running at 1500 rev/min. If the torque on
the shaft driving the generator is 12 Nm.
Determine: (a) the efficiency of the generator.
(b) the power loss in the generator.
Solution:
The efficiency of the generator =
๐‘œ๐‘ข๐‘ก๐‘๐‘ข๐‘ก ๐‘๐‘œ๐‘ค๐‘’๐‘Ÿ
∗ 100 %
๐‘–๐‘›๐‘๐‘ข๐‘ก ๐‘๐‘œ๐‘ค๐‘’๐‘Ÿ
The output power is the electrical output, i.e. V I [Watts]. The input power to a generator is
the mechanical power in the shaft driving the generator, i.e. T ω or T (2πn) [Watts], where T
is the torque in Nm and n is speed of rotation in rev/s. Hence, for a generator
๐‘’๐‘“๐‘“๐‘–๐‘๐‘–๐‘’๐‘›๐‘๐‘ฆ,
๐‘–. ๐‘’.
๐œ‚=
๐œ‚=
๐‘‰∗๐ผ
∗ 100 %
๐‘‡ (2๐œ‹๐‘›)
(100) (15)
∗ 100 % = 79.6 %
1500
(12) (2๐œ‹) (
60 )
๐‘‡โ„Ž๐‘’ ๐‘–๐‘›๐‘๐‘ข๐‘ก ๐‘๐‘œ๐‘ค๐‘’๐‘Ÿ = ๐‘œ๐‘ข๐‘ก๐‘๐‘ข๐‘ก ๐‘๐‘œ๐‘ค๐‘’๐‘Ÿ + ๐‘™๐‘œ๐‘ ๐‘ ๐‘’๐‘ 
๐ป๐‘’๐‘›๐‘๐‘’, ๐‘‡ (2๐œ‹๐‘›) = (๐‘‰ ∗ ๐ผ) + ๐‘™๐‘œ๐‘ ๐‘ ๐‘’๐‘ 
๐‘–. ๐‘’.
๐‘™๐‘œ๐‘ ๐‘ ๐‘’๐‘  = ๐‘‡ (2๐œ‹๐‘›) − (๐‘‰ ∗ ๐ผ)
๐ฟ๐‘œ๐‘ ๐‘ ๐‘’๐‘  = 12 (2 ๐œ‹
1500
) − (100 ∗ 15)
60
๐‘–. ๐‘’. ๐‘๐‘œ๐‘ค๐‘’๐‘Ÿ ๐‘™๐‘œ๐‘ ๐‘ ๐‘’๐‘  = 1885 ๐‘Š − 1500 ๐‘Š = 385 ๐‘Š
10
Problem 20
A 240V shunt motor takes a total current of 30 A. If the field winding resistance Rf =150 โ„ฆ
and the armature resistance Ra = 0.4 โ„ฆ.
Determine: (a) the current in the armature.
(b) the back e.m.f.
Solution:
๐น๐‘–๐‘’๐‘™๐‘‘ ๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก, ๐ผ๐‘“ =
๐‘‰
๐‘…๐‘“
=
240 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
150 Ω
= 1.6 ๐ด
๐‘†๐‘ข๐‘๐‘๐‘™๐‘ฆ ๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก ๐ผ = ๐ผ๐‘“ + ๐ผ๐‘Ž
๐ป๐‘’๐‘›๐‘๐‘’ ๐‘Ž๐‘Ÿ๐‘š๐‘Ž๐‘ก๐‘ข๐‘Ÿ๐‘’ ๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก, ๐ผ๐‘Ž = ๐ผ − ๐ผ๐‘“ = 30 ๐ด − 1.6 ๐ด = 28.4 ๐ด
๐ต๐‘Ž๐‘๐‘˜ ๐‘’. ๐‘š. ๐‘“.
๐ธ = ๐‘‰ − ๐ผ๐‘Ž ๐‘…๐‘Ž
๐ธ = 240 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  − (28.4 ๐ด) (0.4 Ω) = 228.64 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
Problem 21
A 200V, d.c. shunt-wound motor has an armature resistance of 0.4 โ„ฆ and at a certain load has an
armature current of 30A and runs at 1350 rev/min. If the load on the shaft of the motor is
increased so that the armature current increases to 45A, determine the speed of the motor,
assuming the flux remains constant.
Solution:
๐‘‡โ„Ž๐‘’ ๐‘Ÿ๐‘’๐‘™๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘ โ„Ž๐‘–๐‘ ๐ธ ∝ ๐œ™ ๐‘› ๐‘Ž๐‘๐‘๐‘™๐‘–๐‘’๐‘  ๐‘ก๐‘œ ๐‘๐‘œ๐‘กโ„Ž ๐‘”๐‘’๐‘›๐‘’๐‘Ÿ๐‘Ž๐‘ก๐‘œ๐‘Ÿ๐‘  ๐‘Ž๐‘›๐‘‘ ๐‘š๐‘œ๐‘ก๐‘œ๐‘Ÿ๐‘ . ๐น๐‘œ๐‘Ÿ ๐‘Ž ๐‘š๐‘œ๐‘ก๐‘œ๐‘Ÿ๐‘ ,
๐ธ = ๐‘‰ − ๐ผ๐‘Ž ๐‘…๐‘Ž
๐ธ1 = 200 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  − (30 ๐ด) (0.4 Ω) = 188 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
๐ธ2 = 200 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  − (45 ๐ด) (0.4 Ω) = 182 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
๐‘‡โ„Ž๐‘’ ๐‘Ÿ๐‘’๐‘™๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘ โ„Ž๐‘–๐‘,
๐ธ1
๐œ™1 ๐‘›1
=
๐ธ2
๐œ™ 2 ๐‘›2
๐‘Ž๐‘๐‘๐‘™๐‘–๐‘’๐‘  ๐‘ก๐‘œ ๐‘๐‘œ๐‘กโ„Ž ๐‘”๐‘’๐‘›๐‘’๐‘Ÿ๐‘Ž๐‘ก๐‘œ๐‘Ÿ๐‘  ๐‘Ž๐‘›๐‘‘ ๐‘š๐‘œ๐‘ก๐‘œ๐‘Ÿ๐‘ . ๐‘†๐‘–๐‘›๐‘๐‘’ ๐‘กโ„Ž๐‘’ ๐‘“๐‘™๐‘ข๐‘ฅ ๐‘–๐‘  ๐‘๐‘œ๐‘›๐‘ ๐‘ก๐‘Ž๐‘›๐‘ก, ๐œ™1 = ๐œ™2
1350
๐œ™1 ∗ (
188
60 ) ,
๐ป๐‘’๐‘›๐‘๐‘’
=
182
๐œ™1 ∗ ๐‘›2
๐‘–. ๐‘’. ๐‘›2 =
22.5 ∗ 182
๐‘Ÿ๐‘’๐‘ฃ
= 21.78
188
๐‘ 
๐‘‡โ„Ž๐‘ข๐‘  ๐‘กโ„Ž๐‘’ ๐‘ ๐‘๐‘’๐‘’๐‘‘ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘š๐‘œ๐‘ก๐‘œ๐‘Ÿ ๐‘คโ„Ž๐‘’๐‘› ๐‘กโ„Ž๐‘’ ๐‘Ž๐‘Ÿ๐‘š๐‘Ž๐‘ก๐‘ข๐‘Ÿ๐‘’ ๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก ๐‘–๐‘  45๐ด ๐‘–๐‘  21.78 ∗ 60
11
๐‘Ÿ๐‘’๐‘ฃ
๐‘š๐‘–๐‘›
, ๐‘–. ๐‘’. 1307
๐‘Ÿ๐‘’๐‘ฃ
๐‘š๐‘–๐‘›
Problem 22
A 220V, d.c. shunt-wound motor runs at 800 rev/min and the armature current is 30 A. The
armature circuit resistance is 0.4 โ„ฆ.
Determine: (a) the maximum value of armature current if the flux is suddenly reduced by 10 %.
(b) the steady state value of the armature current at the new value of flux, assuming the shaft
torque of the motor remains constant.
Solution:
(a) For a d.c. shunt-wound motor, E = V – Ia Ra. Hence initial generated e.m.f.,
E1 = 220 – 30 × 0.4 = 208V.
The generated e.m.f. is also such that E ∝ n, so at the instant the flux is reduced, the speed
has not had time to change, and E = 208 × 90/100 = 187.2V. Hence, the voltage drop due
to the armature resistance is 220 −187.2, i.e. 32.8V.
The instantaneous value of the current is 32.8/0.4, i.e. 82A. This increase in current is
about three times the initial value and causes an increase in torque, (T ∝ Ia).
The motor accelerates because of the larger torque value until steady state conditions are
reached.
(b) T ∝ Ia and since the torque is constant, ะค1 Ia1 = ะค2 Ia2.
The flux is reduced by 10 %, hence ะค2 = 0.9 ะค1 Thus, ะค1 × 30 = 0.9 ะค1 × Ia2 i.e. the steady
state value of armature current,
๐ผ๐‘Ž2 =
30
1
= 33 ๐ด
0.9
3
12
Problem 23
A series motor has an armature resistance of 0.2 โ„ฆ and a series field resistance of 0.3 โ„ฆ. It is
connected to a 240V supply and at a particular load runs at 24 rev/s when drawing 15 A from
the supply.
(a) Determine the generated e.m.f. at this load.
(b) Calculate the speed of the motor when the load is changed such that the current is increased
to 30 A. Assume that this causes a doubling of the flux.
Solution:
With reference to the Figure, generated e.m.f., E, at initial load,
is given by
๐ธ1 = ๐‘‰ − ๐ผ๐‘Ž (๐‘…๐‘Ž + ๐‘…๐‘“ ) = 240 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  − [(15๐ด) ∗ (0.2 Ω + 0.3Ω)] = 232.5 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
๐‘Šโ„Ž๐‘’๐‘› ๐‘กโ„Ž๐‘’ ๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก ๐‘–๐‘  ๐‘–๐‘›๐‘๐‘Ÿ๐‘’๐‘Ž๐‘ ๐‘’๐‘‘ ๐‘ก๐‘œ 30 ๐ด, ๐‘กโ„Ž๐‘’ ๐‘”๐‘’๐‘›๐‘’๐‘Ÿ๐‘Ž๐‘ก๐‘’๐‘‘ ๐‘’. ๐‘š. ๐‘“. ๐‘–๐‘  ๐‘”๐‘–๐‘ฃ๐‘’๐‘› ๐‘๐‘ฆ:
๐ธ2 = ๐‘‰ − ๐ผ๐‘Ž (๐‘…๐‘Ž + ๐‘…๐‘“ ) = 240 ๐‘‰๐‘œ๐‘™๐‘ก๐‘  − [(30๐ด) ∗ (0.2 Ω + 0.3Ω)] = 225 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
๐‘๐‘œ๐‘ค ๐‘’. ๐‘š. ๐‘“. ๐ธ ∝ Φ๐‘›
๐ธ1
๐œ™1 ๐‘›1
=
๐ธ2
๐œ™ 2 ๐‘›2
๐‘‡โ„Ž๐‘ข๐‘ 
๐‘–. ๐‘’.
232.5
๐œ™1 ∗ (24)
=
(2๐œ™1 ) ∗ (๐‘›2 )
225
๐ป๐‘’๐‘›๐‘๐‘’ ๐‘ ๐‘๐‘’๐‘’๐‘‘ ๐‘œ๐‘“ ๐‘š๐‘œ๐‘ก๐‘œ๐‘Ÿ,
๐‘›2 =
๐‘ ๐‘–๐‘›๐‘๐‘’ ๐œ™2 = 2๐œ™1
(24) (225)
๐‘Ÿ๐‘’๐‘ฃ
= 11.6
(232.5) (2)
๐‘ 
As the current has been increased from 15A to 30A,
the speed has decreased from 24 rev/s to 11.6rev/s.
Its speed/current characteristic is similar to the
Figure.
13
Problem 24
A 320V shunt motor takes a total current of 80 A and runs at 1000 rev/min. If the iron, friction
and windage losses amount to 1.5 kW, the shunt field resistance is 40 โ„ฆ and the armature
resistance is 0.2 โ„ฆ.
- Determine the overall efficiency of the motor.
Solution:
The circuit is shown in the Figure.
๐น๐‘–๐‘’๐‘™๐‘‘ ๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก, ๐ผ๐‘“ =
๐‘‰
๐‘…๐‘“
=
320 ๐‘‰๐‘œ๐‘™๐‘ก๐‘ 
40 Ω
=8๐ด
๐ด๐‘Ÿ๐‘š๐‘Ž๐‘ก๐‘ข๐‘Ÿ๐‘’ ๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก ๐ผ๐‘Ž = ๐ผ − ๐ผ๐‘“ = 80๐ด − 8๐ด = 72๐ด
C = iron, friction and windage losses = 1500 W
Efficiency,
๐‘‰ ∗ ๐ผ − ๐ผ๐‘Ž2 ๐‘…๐‘Ž − ๐ผ๐‘“ ∗ ๐‘‰ − ๐ถ
๐œ‚= (
) ∗ 100 %
๐‘‰∗๐ผ
(320)(80)− (722 )(0.2)− (8)(320)−1500
๐œ‚= (
(320)(80)
) ∗ 100 % =
14
20503
25600
∗ 100 % = 80.1 %
Problem 25
A 250V series motor draws a current of 40 A. The armature resistance is 0.15 โ„ฆ and the field
resistance is 0.05โ„ฆ. - Determine the maximum efficiency of the motor.
Solution:
The circuit is as shown in the Figure. The efficiency,
๐‘‰ ๐ผ − ๐ผ๐‘Ž2 ๐‘…๐‘Ž − ๐ผ๐‘“ ๐‘‰ − ๐ถ
๐œ‚= (
) ∗ 100 %
๐‘‰๐ผ
๐ป๐‘œ๐‘ค๐‘’๐‘ฃ๐‘’๐‘Ÿ ๐‘“๐‘œ๐‘Ÿ ๐‘Ž ๐‘ ๐‘’๐‘Ÿ๐‘–๐‘’๐‘  ๐‘š๐‘œ๐‘ก๐‘œ๐‘Ÿ, ๐ผ๐‘“ = 0 ๐‘Ž๐‘›๐‘‘ ๐‘กโ„Ž๐‘’ ๐ผ๐‘Ž2 ๐‘…๐‘Ž ๐‘™๐‘œ๐‘ ๐‘  ๐‘›๐‘’๐‘’๐‘‘๐‘  ๐‘ก๐‘œ ๐‘๐‘’ ๐ผ 2 (๐‘…๐‘Ž + ๐‘…๐‘“ )
๐‘‰ ๐ผ − ๐ผ 2 (๐‘…๐‘Ž + ๐‘…๐‘“ ) − ๐ถ
๐ป๐‘’๐‘›๐‘๐‘’ ๐‘’๐‘“๐‘“๐‘–๐‘๐‘–๐‘’๐‘›๐‘๐‘ฆ, ๐œ‚ = (
) ∗ 100 %
๐‘‰๐ผ
๐น๐‘œ๐‘Ÿ ๐‘š๐‘Ž๐‘ฅ๐‘–๐‘š๐‘ข๐‘š ๐‘’๐‘“๐‘“๐‘–๐‘๐‘–๐‘’๐‘›๐‘๐‘ฆ ๐ผ 2 (๐‘…๐‘Ž + ๐‘…๐‘“ ) = ๐ถ
๐‘‰ ๐ผ − 2๐ผ 2 (๐‘…๐‘Ž + ๐‘…๐‘“ )
๐ป๐‘’๐‘›๐‘๐‘’ ๐‘’๐‘“๐‘“๐‘–๐‘๐‘–๐‘’๐‘›๐‘๐‘ฆ, ๐œ‚ = (
) ∗ 100 %
๐‘‰๐ผ
(250)(40) − 2(40)2 (0.15 + 0.05)
๐œ‚= (
) ∗ 100 %
(250)(40)
๐œ‚= (
๐œ‚= (
10000 − 640
) ∗ 100 %
10000
9360
) ∗ 100 % = 93.6 %
10000
15
Problem 26
A 200V d.c. motor develops a shaft torque of 15 Nm at 1200 rev/min. If the efficiency is 80%.
- determine the current supplied to the motor.
Solution:
The efficiency of a motor =
๐‘œ๐‘ข๐‘ก๐‘๐‘ข๐‘ก ๐‘๐‘œ๐‘ค๐‘’๐‘Ÿ
∗ 100 %
๐‘–๐‘›๐‘๐‘ข๐‘ก ๐‘๐‘œ๐‘ค๐‘’๐‘Ÿ
The output power of a motor is the power available to do work at its shaft and is given by
T ω or T(2πn) [watts], where T is the torque in Nm and n is the speed of rotation in [rev/s].
The input power is the electrical power in watts supplied to the motor, i.e. VI [watts].
Thus for a motor, efficiency,
๐œ‚=
i.e.
๐‘‡ (2 ๐œ‹ ๐‘›)
๐‘‰๐ผ
∗ 100 %
80 = [
(15) (2 ๐œ‹) (
1200
)
60
(200) (๐ผ)
] (100)
๐‘‡โ„Ž๐‘ข๐‘  ๐‘กโ„Ž๐‘’ ๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก ๐‘ ๐‘ข๐‘๐‘๐‘™๐‘–๐‘’๐‘‘, ๐ผ =
(15) (2๐œ‹) (20) (100)
(200) (80)
= 11.8 ๐ด
Problem 27
A d.c. series motor drives a load at 30 rev/s and takes a current of 10 A when the supply voltage
is 400V. If the total resistance of the motor is 2 โ„ฆ and the iron, friction and windage losses amount
to 300W, determine the efficiency of the motor.
Solution:
๐‘‰ ๐ผ − ๐ผ 2 ๐‘…−๐ถ
๐ธ๐‘“๐‘“๐‘–๐‘๐‘–๐‘’๐‘›๐‘๐‘ฆ, ๐œ‚ = (
๐‘‰๐ผ
) ∗ 100 %
(400)(10)− (10)2 (2)−300
๐ธ๐‘“๐‘“๐‘–๐‘๐‘–๐‘’๐‘›๐‘๐‘ฆ, ๐œ‚ = (
(400) (10)
4000−200−300
๐ธ๐‘“๐‘“๐‘–๐‘๐‘–๐‘’๐‘›๐‘๐‘ฆ, ๐œ‚ = (
4000
) ∗ 100 %
) ∗ 100 % = 87.5 %
16
Problem 28
A 500V shunt motor runs at its normal speed of 10 rev/s when the armature current is 120A.
The armature resistance is 0.2 โ„ฆ.
(a) Determine the speed when the current is 60A and a
resistance of 0.5 โ„ฆ is connected in series with the
armature, the shunt field remaining constant.
(b) Determine the speed when the current is 60A and
the shunt field is reduced to 80 % of its normal value by increasing resistance in the field circuit.
Solution:
With reference to the Figure, back e.m.f. at 120A,
๐ธ1 = ๐‘‰ − ๐ผ๐‘Ž ๐‘…๐‘Ž = 500 ๐‘‰ − [(120๐ด) ∗ (0.2 Ω)] = 476 ๐‘‰
๐‘Šโ„Ž๐‘’๐‘› ๐ผ๐‘Ž = 60 ๐ด, โ‰ซโ‰ซโ‰ซ ๐ธ2 = 500 ๐‘‰ − [(60๐ด) ∗ (0.2 Ω + 0.5Ω)] = 458 ๐‘‰
๐‘๐‘œ๐‘ค
๐ธ1
๐œ™1 ๐‘›1
=
๐ธ2
๐œ™ 2 ๐‘›2
476
๐œ™1 (10)
=
458
๐œ™ 2 ( ๐‘›2 )
๐‘–. ๐‘’.
๐’‡๐’“๐’๐’Ž ๐’˜๐’‰๐’Š๐’„๐’‰, ๐’”๐’‘๐’†๐’†๐’… ๐’๐Ÿ =
๐‘ ๐‘–๐‘›๐‘๐‘’ ∅2 = ∅1
(๐Ÿ๐ŸŽ) (๐Ÿ’๐Ÿ“๐Ÿ–)
๐’“๐’†๐’—
= ๐Ÿ—. ๐Ÿ”๐Ÿ [
]
(๐Ÿ’๐Ÿ•๐Ÿ”)
๐’”
๐‘ฉ๐’‚๐’„๐’Œ ๐’†. ๐’Ž. ๐’‡. ๐’˜๐’‰๐’†๐’ ๐‘ฐ๐’‚ = ๐Ÿ”๐ŸŽ ๐‘จ,
๐ธ2 = 500 ๐‘‰ − [(60๐ด) ∗ (0.2 Ω)] = 488 ๐‘‰
๐‘๐‘œ๐‘ค
๐ธ1
๐œ™1 ๐‘›1
=
๐ธ2
๐œ™2 ๐‘›2
๐‘–. ๐‘’.
476
๐œ™1 (10)
=
488
0.8 ๐œ™1 (๐‘›3 )
๐’‡๐’“๐’๐’Ž ๐’˜๐’‰๐’Š๐’„๐’‰, ๐’”๐’‘๐’†๐’†๐’… ๐’๐Ÿ‘ =
๐‘ ๐‘–๐‘›๐‘๐‘’ ∅2 = 0.8 ∅1
(๐Ÿ๐ŸŽ) (๐Ÿ’๐Ÿ–๐Ÿ–)
๐’“๐’†๐’—
= ๐Ÿ๐Ÿ. ๐Ÿ–๐Ÿ [
]
(๐ŸŽ. ๐Ÿ–)(๐Ÿ’๐Ÿ•๐Ÿ”)
๐’”
17
Problem 29
On full-load a 300 V series motor takes 90 A and runs at 15 rev/s. The armature resistance is
0.1โ„ฆ and the series winding resistance is 50 mโ„ฆ.
- Determine the speed when developing full load torque but with a 0.2 โ„ฆ diverter in parallel
with the field winding. (Assume that the flux is proportional to the field current.)
Solution:
๐ด๐‘ก 300 ๐‘‰, ๐‘’. ๐‘š. ๐‘“. ๐ธ1 = ๐‘‰ − ๐ผ๐‘…
= ๐‘‰ − ๐ผ(๐‘…๐‘Ž + ๐‘…๐‘ ๐‘’ )
= 300 − (90)(0.1 + 0.05)
= 300 − 13.5 = 286.5 ๐‘ฃ๐‘œ๐‘™๐‘ก๐‘ 
๐‘Š๐‘–๐‘กโ„Ž ๐‘กโ„Ž๐‘’ 0.2 Ω ๐‘‘๐‘Ÿ๐‘–๐‘ฃ๐‘’๐‘Ÿ ๐‘–๐‘› ๐‘๐‘Ž๐‘Ÿ๐‘Ž๐‘™๐‘™๐‘’๐‘™ ๐‘ค๐‘–๐‘กโ„Ž ๐‘…๐‘ ๐‘’
(0.2)(0.05)
(0.2)(0.05)
=
= 0.04 Ω
(0.2) + (0.05)
(0.25)
0.2
๐ต๐‘ฆ ๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก ๐‘‘๐‘–๐‘ฃ๐‘–๐‘ ๐‘–๐‘œ๐‘›, ๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก ๐ผ1 = (
) ๐ผ = 0.8 ๐ผ
0.2 + 0.05
๐‘กโ„Ž๐‘’ ๐‘’๐‘ž๐‘ข๐‘–๐‘ฃ๐‘Ž๐‘™๐‘’๐‘›๐‘ก ๐‘Ÿ๐‘’๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’, ๐‘… =
๐‘‡๐‘œ๐‘Ÿ๐‘ž๐‘ข๐‘’, ๐‘‡ ๐›ผ ๐ผ๐‘Ž ∅ ๐‘Ž๐‘›๐‘‘ ๐‘“๐‘œ๐‘Ÿ ๐‘“๐‘ข๐‘™๐‘™ ๐‘™๐‘œ๐‘Ž๐‘‘ ๐‘ก๐‘œ๐‘Ÿ๐‘ž๐‘ข๐‘’, ๐ผ๐‘Ž1 ∅1 = ๐ผ๐‘Ž2 ∅2
๐‘ ๐‘–๐‘›๐‘๐‘’ ๐‘“๐‘™๐‘ข๐‘ฅ ๐‘–๐‘  ๐‘๐‘Ÿ๐‘œ๐‘๐‘œ๐‘Ÿ๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘ก๐‘œ ๐‘“๐‘–๐‘’๐‘™๐‘‘ ๐‘๐‘ข๐‘Ÿ๐‘Ÿ๐‘’๐‘›๐‘ก ∅1 ∝ ๐ผ๐‘Ž1 ๐‘Ž๐‘›๐‘‘ ∅2 ∝ 0.8 ๐ผ๐‘Ž2
๐‘‡โ„Ž๐‘’๐‘› (90)(90) = (๐ผ๐‘Ž2 )(0.8 ๐ผ๐‘Ž2 )
๐‘“๐‘Ÿ๐‘œ๐‘š
2
๐‘คโ„Ž๐‘–๐‘โ„Ž, ๐ผ๐‘Ž2
(90)2
90
=
๐‘Ž๐‘›๐‘‘ ๐ผ๐‘Ž2 =
= 100.62 ๐ด
0.8
√(0.8)
๐ป๐‘’๐‘›๐‘๐‘’ ๐‘’. ๐‘š. ๐‘“. ๐ธ2 = ๐‘‰ − ๐ผ๐‘Ž2 (๐‘…๐‘Ž + ๐‘…)
= 300 − (100.62)(0.1 + 0.04)
= 300 − (100.62)(0.14)
= 300 − 14.087 = 285.9 ๐‘ฃ๐‘œ๐‘™๐‘ก๐‘ 
๐‘๐‘œ๐‘ค ๐‘’. ๐‘š. ๐‘“. ๐ธ ∝ ∅๐‘› ๐‘“๐‘Ÿ๐‘œ๐‘š ๐‘คโ„Ž๐‘–๐‘โ„Ž,
๐ป๐‘’๐‘›๐‘๐‘’
๐ธ1
∅1 ๐‘›1
๐ผ๐‘Ž1 ๐‘›1
=
=
๐ธ2
∅2 ๐‘›2
0.8 ๐ผ๐‘Ž2 ๐‘›2
(286.5)
(90)(15)
=
285.9
(0.8)(100.62) ๐‘›2
(285.9)(90)(15)
= 16.74 ๐‘Ÿ๐‘’๐‘ฃ/๐‘ 
(286.5)(0.8)(100.62)
๐‘Ÿ๐‘’๐‘ฃ
๐‘‡โ„Ž๐‘ข๐‘  ๐‘กโ„Ž๐‘’ ๐‘ ๐‘๐‘’๐‘’๐‘‘ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘š๐‘œ๐‘ก๐‘œ๐‘Ÿ โ„Ž๐‘Ž๐‘  ๐‘–๐‘›๐‘๐‘Ÿ๐‘’๐‘Ž๐‘ ๐‘’๐‘‘ ๐‘“๐‘Ÿ๐‘œ๐‘š 15
๐‘ ๐‘’๐‘
๐‘Ž๐‘›๐‘‘ ๐‘›๐‘’๐‘ค ๐‘ ๐‘๐‘’๐‘’๐‘‘, ๐‘›2 =
๐‘–. ๐‘’. 900 ๐‘Ÿ๐‘’๐‘ฃ/๐‘š๐‘–๐‘›
๐‘ก๐‘œ 16.74 ๐‘Ÿ๐‘’๐‘ฃ/๐‘ ๐‘’๐‘ i.e., 1004 rev/min by inserting a 0.2 โ„ฆ diverter resistance in parallel with the
series winding.
18
Problem 30
A series motor runs at 800 rev/min when the voltage is 400 V and the current is 25 A. The
armature resistance is 0.4 โ„ฆ and the series field resistance is 0.2 โ„ฆ.
- Determine the resistance to be connected in series to reduce the
speed to 600 rev/min with the same current.
Solution:
With reference to the Figure, at 800 rev/min,
๐’†. ๐’Ž. ๐’‡. , ๐‘ฌ๐Ÿ = ๐‘ฝ − ๐‘ฐ (๐‘น๐’‚ + ๐‘น๐’”๐’† ) = ๐Ÿ’๐ŸŽ๐ŸŽ ๐‘ฝ − (๐Ÿ๐Ÿ“๐‘จ) ∗ (๐ŸŽ. ๐Ÿ’๐›€ + ๐ŸŽ. ๐Ÿ๐›€) = ๐Ÿ‘๐Ÿ–๐Ÿ“ ๐‘ฝ
At 600 rev/min, since the current is unchanged, the flux is unchanged.
๐‘‡โ„Ž๐‘ข๐‘  ๐ธ ∝ ∅ ๐‘›, ๐‘œ๐‘Ÿ ๐ธ ∝ ๐‘›, ๐‘Ž๐‘›๐‘‘
๐ธ1
๐‘›1
=
๐ธ2
๐‘›2
๐’‡๐’“๐’๐’Ž ๐’˜๐’‰๐’Š๐’„๐’‰, ๐‘ฌ๐Ÿ =
๐‘Ž๐‘›๐‘‘
๐‘ฏ๐’†๐’๐’„๐’†
๐Ÿ‘๐Ÿ–๐Ÿ“
๐Ÿ–๐ŸŽ๐ŸŽ
=
๐‘ฌ๐Ÿ
๐Ÿ”๐ŸŽ๐ŸŽ
(๐Ÿ‘๐Ÿ–๐Ÿ“) (๐Ÿ”๐ŸŽ๐ŸŽ)
= ๐Ÿ๐Ÿ–๐Ÿ–. ๐Ÿ•๐Ÿ“ ๐‘ฝ
(๐Ÿ–๐ŸŽ๐ŸŽ)
๐ธ2 = ๐‘‰ − ๐ผ (๐‘…๐‘Ž + ๐‘…๐‘ ๐‘’ + ๐‘…)
๐ป๐‘’๐‘›๐‘๐‘’ 288.75 = 400 ๐‘‰ − [(25๐ด) ∗ (0.4 Ω + 0.2Ω + R)]
๐‘…๐‘’๐‘Ž๐‘Ÿ๐‘Ÿ๐‘Ž๐‘›๐‘”๐‘–๐‘›๐‘” ๐‘”๐‘–๐‘ฃ๐‘’๐‘ : 0.6 + ๐‘… =
400 − 288.75
= 4.45
25
๐’‡๐’“๐’๐’Ž ๐’˜๐’‰๐’Š๐’„๐’‰, ๐’†๐’™๐’•๐’“๐’‚ ๐’”๐’†๐’“๐’Š๐’†๐’” ๐’“๐’†๐’”๐’Š๐’”๐’•๐’‚๐’๐’„๐’†, ๐‘น = ๐Ÿ’. ๐Ÿ’๐Ÿ“ − ๐ŸŽ. ๐Ÿ” = ๐Ÿ‘. ๐Ÿ–๐Ÿ“ ๐›€
Thus, the addition of a series resistance of 3.85 โ„ฆ has reduced the speed from 800 rev/min to
600 rev/min.
19
Appendix
20
[๐‘‰ ]
๐ธ = ๐‘˜๐‘Ž ∅ ๐œ”๐‘š
๐‘‡=
๐‘ƒ
๐ธ ๐ผ๐‘Ž
=
= ๐‘˜๐‘Ž ∅ ๐ผ๐‘Ž
๐œ”๐‘š
๐œ”๐‘š
[๐‘. ๐‘š]
The principal losses of machines are:
(i) Copper loss, due to I 2R heat losses in the armature and field windings.
(ii) Iron (or core) loss, due to hysteresis and eddy current losses in the armature.
This loss can be reduced by constructing the armature of silicon steel
laminations having a high resistivity and low hysteresis loss. At constant speed,
the iron loss is assumed constant.
(iii) Friction and windage losses, due to bearing and brush contact friction and
losses due to air resistance against moving parts (called windage).
At constant speed, these losses are assumed to be constant.
(iv) Brush contact loss between the brushes and commutator.
This loss is approximately proportional to the load current.
The total losses of a machine can be quite significant and operating efficiencies
of between 80% and 90% are common.
Efficiency of a d.c. generator
The efficiency of an electrical machine is the ratio of the output power to the input
power and is usually expressed as a percentage. The Greek letter ‘η’ (eta) is used to
signify efficiency and since the units are power/power, then efficiency has no units.
Thus
21
If the total resistance of the armature circuit (including brush contact resistance) is
Ra, then the total loss in the armature circuit is (๐‘ฐ๐Ÿ๐’‚ ∗ ๐‘น๐’‚ )
If the terminal voltage is V and the current in the shunt circuit is If, then the loss
in the shunt circuit is (If *V).
If the sum of the iron, friction and windage losses is C then the total losses is given
by:
If the output current is I, then the output power is VI.
Total input power = ๐‘‰ ๐ผ + ๐ผ๐‘Ž2 ๐‘…๐‘Ž + ๐ผ๐‘“ ๐‘‰ + ๐ถ. Hence
The efficiency of a generator is a maximum when the load is such that:
i.e. when the variable loss = the constant loss
22
The efficiency of a d.c. motor
the efficiency of a d.c. machine is given by:
Also, the total losses = ๐ผ๐‘Ž2 ๐‘…๐‘Ž + ๐ผ๐‘“ ๐‘‰ + ๐ถ (for a shunt motor) where C is the sum
of the iron, friction and windage losses.
For a motor, the input power = VI
and the output power = VI – losses
Hence
The efficiency of a motor is a maximum when the load is such that:
Also, the total losses = ๐ผ๐‘Ž2 (๐‘…๐‘Ž + ๐‘…๐‘“ ) + ๐ถ (for a series motor) where C is the sum
of the iron, friction and windage losses.
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