Long shunt compound generator emf, 80 A at 230 V

In a long shunt compound generator the shunt field is connected across the load terminals, so the series field carries the full armature current.

Given: $I_L = 80$ A, $V = 230$ V, $R_{sh} = 100\ \Omega$, $R_{se} = 0.04\ \Omega$, $R_a = 0.2\ \Omega$.

       Ia        Rse
  +--->---[~~~~]---+---------+----o +
  |                |         |
 (E,Ra)          [Rsh]     Load
  |                |         |
  +----------------+---------+----o -

Step 1: Shunt field current

$$I_{sh} = \frac{V}{R_{sh}} = \frac{230}{100} = 2.30\ \text{A}$$

Step 2: Armature current

$$I_a = I_{se} = I_L + I_{sh} = 80 + 2.30 = 82.30\ \text{A}$$

Step 3: Generated emf

$$E = V + I_a(R_a + R_{se}) = 230 + 82.30(0.2 + 0.04) = 230 + 19.752 = 249.75\ \text{V}$$

Answer: emf generated by the armature $E = 249.75\ \text{V}$.