Topic: Electrical Machines
A DC shunt generator produces 450A at 230V. The resistance of shunt field and armature are 50 ohms and 0.025 ohm respectively. The armature voltage drop will be
- A. 11.36 V
- B. 22.7 V
- C. 31.6 V
- D. 38.4 V
Answer: A. 11.36 V
Explanation: For a DC shunt generator, the terminal voltage $V = 230V$, Load current $I_L = 450A$.
Shunt field current $I_{sh} = \frac{V}{R_{sh}} = \frac{230}{50} = 4.6A$.
Armature current $I_a = I_L + I_{sh} = 450 + 4.6 = 454.6A$.
Armature voltage drop = $I_a \times R_a = 454.6 \times 0.025 = 11.365V$.