Question

A 100-kW, 250-V DC shunt generator has an armature resistance of 0.05Ω. With the generator operating at rated voltage, determine the induced voltage at full load.

Answer 1

Answer on Question #64392, Engineering / Electrical Engineering

A 100-kW, 250-V DC shunt generator has an armature resistance of $0.05\Omega$ and a field circuit resistance of $60\Omega$. With the generator operating at rated voltage, determine the induced voltage at full load.

Solution:

In a shunt generator, the field winding is connected in parallel with the armature winding so that terminal voltage of the generator is applied across it.

At full load, the terminal voltage

$$V = E _ {g} - I _ {a} R _ {a}$$

Shunt field current,

$$I _ {s h} = \frac {V}{R _ {s h}} = \frac {2 5 0 V}{6 0 \Omega} = 4. 1 7 \mathrm {A}$$

Load current,

$$I _ {L} = \frac {P}{V} = \frac {1 0 0 , 0 0 0 W}{2 5 0 V} = 4 0 0 \mathrm {A}$$

Armature current,

$$I _ {a} = I _ {L} + I _ {s h} = 4 0 0 + 4. 1 7 = 4 0 4. 1 7 \mathrm {A}$$

Induced voltage,

$$E _ {g} = V + I _ {a} R _ {a} = 2 5 0 V + (4 0 4. 1 7 A) \times (0. 0 5 \Omega) = 2 7 0. 2 \mathrm {V}$$

Answer: 270.2 V

Answer provided by AssignmentExpert.com