Short shunt compound generator, 100 A at 250 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: $I_L = 100$ A, $V = 250$ V, $R_{sh} = 130\ \Omega$, $R_{se} = 0.1\ \Omega$, $R_a = 0.1\ \Omega$.

Brush drop $= 2$ V (1 V per brush, 2 brushes).

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

Step 1: Voltage across the shunt field

$$V_{sh} = V + I_LR_{se} = 250 + 100(0.1) = 260.00\ \text{V}$$

Step 2: Currents

$$I_{sh} = \frac{V_{sh}}{R_{sh}} = \frac{260.00}{130} = 2.000\ \text{A}, \qquad I_a = I_L + I_{sh} = 100 + 2.000 = 102.000\ \text{A}$$

Step 3: Generated emf

$$E = V_{sh} + I_aR_a + 2 = 260.00 + (102.000)(0.1) + 2 = 272.20\ \text{V}$$

Answer: generated emf $E = 272.20\ \text{V}$ and armature current $I_a = 102.000\ \text{A}$.