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

Explanation:

Step 1: Understanding the parameters

• Generated Current (I_G): The total current generated by the DC shunt generator is given as 450A.
• Terminal Voltage (V_T): The terminal voltage of the generator is given as 230V.
• Shunt Field Resistance (R_sh): The resistance of the shunt field is 50 ohms.
• Armature Resistance (R_a): The resistance of the armature is 0.025 ohms.

Step 2: Calculating the shunt field current (I_sh)

The shunt field current can be calculated using Ohm's law:

I_sh = V_T / R_sh

Substituting the known values:

I_sh = 230 / 50 = 4.6 A

Step 3: Calculating the armature current (I_a)

The armature current can be determined as the difference between the total generated current (I_G) and the shunt field current (I_sh):

I_a = I_G - I_sh

Substituting the known values:

I_a = 450 - 4.6 = 445.4 A

Step 4: Calculating the armature voltage drop (V_a)

The armature voltage drop is given by Ohm's law:

V_a = I_a × R_a

Substituting the known values:

V_a = 445.4 × 0.025 = 11.36 V

Conclusion:

The armature voltage drop is 11.36V, making Option 1 the correct answer.