Compound generator EMF in long and short shunt

Given: $I_L = 120$ A, $V = 120$ V, $R_{sh} = 30\ \Omega$, $R_{se} = 0.05\ \Omega$, $R_a = 0.1\ \Omega$. Brush drop is neglected.

(i) Long-shunt connection

The series field is in series with the armature; the shunt field is across the load terminals.

       Ia    Rse              IL = 120 A
   +--->---/\/\/--+------+--->---o +
   |              |      |
 ( E )          [Rsh]  load     V = 120 V
   Ra             |      |
   +--------------+------+-------o -

$$\begin{aligned} I_{sh} &= \frac{V}{R_{sh}} = \frac{120}{30} = 4\ \text{A} \\ I_a &= I_L + I_{sh} = 120 + 4 = 124\ \text{A} \\ E_g &= V + I_a(R_a + R_{se}) = 120 + 124 \times (0.1 + 0.05) \\ &= 120 + 18.6 = 138.6\ \text{V} \end{aligned}$$

(ii) Short-shunt connection

The shunt field is across the armature; the series field carries the load current.

       Ia            Rse       IL = 120 A
   +--->---+-------/\/\/---+--->---o +
   |       |               |
 ( E )   [Rsh]           load    V = 120 V
   Ra      |               |
   +-------+---------------+-------o -

$$\begin{aligned} V_{sh} &= V + I_LR_{se} = 120 + 120 \times 0.05 = 126\ \text{V} \\ I_{sh} &= \frac{126}{30} = 4.2\ \text{A} \\ I_a &= I_L + I_{sh} = 120 + 4.2 = 124.2\ \text{A} \\ E_g &= V_{sh} + I_aR_a = 126 + 124.2 \times 0.1 \\ &= 126 + 12.42 = 138.42\ \text{V} \end{aligned}$$

Answer: (i) Long shunt: $E_g =$ 138.6 V; (ii) Short shunt: $E_g =$ 138.42 V.