A short shunt compound generator delivers 30A to the load at terminal voltage of 220V. The resistances of armature, series and shunt field are 0.05 , 0.03 and 250 . Find out the current flowing through the armature and the voltage across the armature. The voltage drop per brush is 1V.
Hello!
(https://drive.google.com/file/d/15IfznjMnKxnxrZXoO6ujigj2Q2SvOjky/view?usp=sharing)
Please consider circuit diagram uploaded in the above google drive link.
i2 = 30 A (given)
i2*R = 220 V
Let E = required voltage across armature, and
Let i1 = current flowing through the armature
We have to find E and i1
i1 = i2 + i3 (Equation 1)
The contact drop per brush is 1 V (given)
Therefore, for two brushes, contact drop will be 2 * 1 V = 2 V
Applying Kirchhoff’s Voltage Law (KVL) in loop 1, we get
0.03*i2 + i2*R + 0.05*i1 + 2 = E
Substituting the value of i2 and i2*R, we get:
0.03*30 + 220 + 0.05*i1 + 2 = E
222.9 + 0.05*i1 = E
Applying Kirchhoff’s Voltage Law (KVL) in loop 2, we get:
0.05*i1 + 250*i3 = 0
0.05*i1 + 250*(i1-i2) = 0
0.05*i1 + 250*i1 – 250*i2 = 0
Substituting the value of i2, we get:
250.05*i1 = 7500
Therefore, i1 = 7500/250.05 = 29.99
Current flowing through the armature, i1 = 29.99 A (Answer)
Now, substituting the value of i1 in equation 2, we get the value of E as:
222.9 + 0.05*(7500/250.05) = E
E = 222.9 + 1.49
Voltage across the armature, E = 224.399 ~ 225 V (Answer)
Hope this helps.
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