A 500 V shunt motor runs at its normal speed of 250 RPM when the armature current is 200 A. The resi

Question

A 500 V shunt motor runs at its normal speed of 250 RPM when the armature current is 200 A. The resistance of armature is 0.12 Ω. Calculate the speed when resistance is inserted in the field reducing the shunt field to 80% of normal value and the armature current is 100A.

A

420 RPM

B

540 RPM

C

320 RPM

D

1500 RPM

Examoogle · Accepted Answer

320 RPM

Explanation: Given: Supply Voltage V = 500V, Initial Speed N1 = 250 RPM, Armature Resistance Ra = 0.12 Ω, Initial Armature Current Ia1 = 200A, Final Armature Current Ia2 = 100A, Final Flux φ2 = 0.8 φ1.

Formula Used: $$E = V - I_a R_a$$, $$N \propto \frac{E}{\phi}$$

Steps: Calculate initial back EMF (E1): Using the formula for back EMF for a DC motor: $E_1 = V - I_{a1} R_a$.. Calculate final back EMF (E2): Calculate the back EMF at the new armature current: $E_2 = V - I_{a2} R_a$.. Relate speed to back EMF and flux: Since $N \propto \frac{E}{\phi}$, we use the ratio: $\frac{N_2}{N_1} = \frac{E_2}{E_1} 	imes \frac{\phi_1}{\phi_2}$. Given $\phi_2 = 0.8 \phi_1$, the ratio $\frac{\phi_1}{\phi_2} = \frac{1}{0.8} = 1.25$.. Final Calculation: Evaluating the expression to find the new speed $N_2$.

Answer: C is correct because the calculated speed after the flux reduction and current change is approximately 320 RPM.

Exam Tip: In DC shunt motors, remember that reducing field flux always leads to an increase in speed, which is a fundamental concept in field weakening control techniques.

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