Two D.C. shunt generators, each with armature resistance of 0.02 ohm and field resistance of 50 ohm run in parallel and supply a total current of 1000 amperes to the load circuit. If their e.m.fs. are 270 V and 265 V, their bus bar voltage will be
A. 270 V
B. 267.5 V
C. 265 V
D. 257.5 V
Answer: Option E
Solution (By Examveda Team)
The two D.C. shunt generators are running in parallel, so they share the total load current of 1000 A based on their induced e.m.fs. and internal characteristics.Let I₁ and I₂ be the currents supplied by generators 1 and 2 respectively.
For generator 1 (E₁=270 V): Bus bar voltage V = E₁ - I₁ × 0.02
For generator 2 (E₂=265 V): Bus bar voltage V = E₂ - I₂ × 0.02
Since both are connected to the same bus bars, their bus bar voltages must be equal, so:
E₁ - 0.02 × I₁ = E₂ - 0.02 × I₂
Rearranging gives: 0.02(I₂ - I₁) = E₂ - E₁ = 265 - 270 = -5
So: I₂ - I₁ = -5 / 0.02 = -250 ⇒ I₂ = I₁ - 250
Also, total current constraint gives: I₁ + I₂ = 1000
Substituting I₂: I₁ + (I₁ - 250) = 1000 ⇒ 2I₁ - 250 = 1000 ⇒ 2I₁ = 1250 ⇒ I₁=625 A
Then I₂=625-250=375 A
Finally, bus bar voltage V can be found using either generator (using generator 1):
V = E₁ - I₁ × 0.02 = 270 - (625×0.02) = 270 -12.5= 257.5 V
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Related Questions on D.C. Generators
A cumulatively compounded long shunt generator when operating as a motor would be
A. Cumulatively compounded long shunt
B. Differentially compounded long shunt
C. Cumulatively compounded short shunt
D. Differentially compounded short shunt
A. Demagnetization only
B. Cross magnetization as well as magnetization
C. Cross-magnetization as well as demagnetizing
D. Cross magnetization only
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Ans=257.4 (d) .similar math A Textbook of Electrical Technology ( Volume-ii) by B.L. THERAJA & A.K. THERAJA
The right ans is 257.5v
Bus Bar Voltage=
∑(Individual Field Resistances)
∑(Individual E.M.F.s×Individual Field Resistances)+Total Load Current×Total Armature Resistance
Let's calculate:
Bus Bar Voltage
=
(
270
×
50
)
+
(
265
×
50
)
+
(
1000
×
0.02
)
50
+
50
Bus Bar Voltage=
50+50
(270×50)+(265×50)+(1000×0.02)
Bus Bar Voltage
=
13500
+
13250
+
20
100
Bus Bar Voltage=
100
13500+13250+20
Bus Bar Voltage
=
26770
100
Bus Bar Voltage=
100
26770
Bus Bar Voltage
=
267.7
V
Bus Bar Voltage=267.7V
I1+I2=1000A
I1=Eg1-v/Ra
I2=Eg2-v/Ra
I1+I2=270-v/Ra+265-v/Ra
Solve it than we find
2v=535
V=535/2
V =267.5v
D is the correct answer
I1+I2=1000A
I1=Eg1-V/Ra
I2= Eg2-V/Ra
I1+I2= 270-V/Ra +265-V/Ra
Solve it than we find
2V=515
v= 257.5
In DC generators brushes are used for
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Please solve this problem
The given answer is wrong.The full solution is little bit lengthy,so we should try it using options.Since armature resistance is identical in both generators,the generator with high emf should supply more current (i.e. more than 500A)to the load.So the drop on the armature resistance should be more than 500A *0.02 =10v.So the terminal voltage must be less than 270-10=260V.By looking at the option the only possible answer is D.
the avareage
Please solve this problem
Please solve this problem