Short shunt compound generator emf, 80 A at 220 V

In a short shunt compound generator the shunt field is connected across the armature only, so the series field carries the load current $I_L$ and the shunt field sees $V + I_LR_{se}$.

Given: $I_L = 80$ A, $V = 220$ V, $R_{sh} = 100\ \Omega$, $R_{se} = 0.05\ \Omega$, $R_a = 0.1\ \Omega$.

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

Step 1: Voltage across the shunt field

$$V_{sh} = V + I_LR_{se} = 220 + 80(0.05) = 224.00\ \text{V}$$

Step 2: Currents

$$I_{sh} = \frac{V_{sh}}{R_{sh}} = \frac{224.00}{100} = 2.240\ \text{A}, \qquad I_a = I_L + I_{sh} = 80 + 2.240 = 82.240\ \text{A}$$

Step 3: Generated emf

$$E = V_{sh} + I_aR_a = 224.00 + (82.240)(0.1) = 232.22\ \text{V}$$

Answer: generated emf $E = 232.22\ \text{V}$.