Question

Swinburne’s test measures the following on a 200 V, 15 kW dc shunt motor:
Input current = 7 A; Field current = 2 A. Estimate field copper loss and iron \& frictional losses. Assume \( R_a = 0.04 \Omega \)

• A. 400 W, 998 W respectively
• B. 999 W, 400 W respectively
• C. 400 W, 999 W respectively
• D. 998 W, 400 W respectively

Hint:

Total loss = Input power – Field loss – Armature loss.

Answer 1

The Correct Option is C

Field copper loss = \( V \times I_f = 200 \times 2 = 400 \, \text{W} \)
Armature current = \( 7 - 2 = 5 \, \text{A} \Rightarrow \) Armature copper loss = \( I_a^2 R_a = 25 \times 0.04 = 1 \, \text{W} \)
Total loss = \( 200 \times 7 = 1400 \, \text{W} \)
Iron + friction loss = \( 1400 - 400 - 1 = 999 \, \text{W} \)