Question 16.18: In a 200 V, 20 A, DC motor, the magnetic flux suddenly incre......
In a 200 V, 20 A, DC motor, the magnetic flux suddenly increases to about 10%. Calculate the current that this machine will deliver to the electrical grid. Herein, assume that the magnetic circuit of the machine is linear and the armature circuit resistance is 0.1 Ω.
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1) 178 A
2) 79 A
3) 169 A
4) 99 A
Based on the information given in the problem, we have:
V_t=200\ V (1)
I_t=20\ A (2)
φ_2=1.1φ_1 (3)
φ \propto I_f (4)
R_a=0.1\ \Omega (5)
Figure 16.18 shows the electrical circuit of a shunt DC motor. Applying KVL in the right-hand side mesh of the circuit:
E_{a,1}=V_t-R_aI_{a,1}=200-(0.1\times 20)=198\ V (6)
Now, if the magnetic flux suddenly increases to about 10%, the speed of motor can be assumed constant (ω_2 = ω_1) for a short period of time. The new electromotive force (emf) of the machine can be achieved as follows, since the magnetic circuit of machine is linear:
φ_2=1.1φ_1\overset{E_a\propto φ}{\Longrightarrow } E_{a,2}=1.1E_{a,1}=1.1\times 198=217.8 (7)
As can be seen, E_{a,2}>V_t. Therefore, the machine is now behaving like a generator.
Applying KVL in the right-hand side mesh of the circuit:
E_{a,2}=V_t-R_aI_{a,2} (8)
\Rightarrow 217.8=200-(0.1I_{a_2})\Rightarrow I_{a,2}=-178\ A\Rightarrow \pmb{|I_{a,2}|=178\ A}
The negative sign of the current indicates that the current direction has changed and shows that the machine is now a generator.
Choice (1) is the answer.
