Short shunt to long shunt compound generator (10 kW, 220 V)

Given: $P = 10$ kW, $V = 220$ V, $R_{sh} = 100\ \Omega$, $R_{se} = 0.3\ \Omega$, $R_a = 0.4\ \Omega$. Brush drop neglected.

Case 1: Short-shunt connection

     +--Ra--+-------Rse-------+
     |      |                 |
    (Eg)   Rsh              Load 220 V
     |      |                 |
     +------+-----------------+

$$\begin{aligned} I_L &= \frac{10000}{220} = 45.455\ \text{A} \\ V_{sh} &= V + I_L R_{se} = 220 + 45.455 \times 0.3 = 233.64\ \text{V} \\ I_{sh} &= \frac{233.64}{100} = 2.336\ \text{A} \\ I_a &= I_L + I_{sh} = 47.791\ \text{A} \\ E_g &= V_{sh} + I_a R_a = 233.64 + 47.791 \times 0.4 \\ &= 233.64 + 19.12 = 252.75\ \text{V} \end{aligned}$$

Case 2: Reconnected as long shunt, same $E_g$ and $I_a$

     +--Ra--Rse--+-----------+
     |           |           |
    (Eg)        Rsh        Load  V
     |           |           |
     +-----------+-----------+

Now the armature current flows through both $R_a$ and $R_{se}$:

$$\begin{aligned} V &= E_g - I_a (R_a + R_{se}) \\ &= 252.75 - 47.791 \times (0.4 + 0.3) \\ &= 252.75 - 33.45 = 219.30\ \text{V} \end{aligned}$$

Check: $I_{sh} = 219.30/100 = 2.193$ A, $I_L = 47.791 - 2.193 = 45.60$ A, which is consistent.

Answer: Case 1: induced emf $E_g \approx 252.75$ V. Case 2: terminal voltage in long shunt $V \approx 219.3$ V (slightly lower, because the series field now carries the larger armature current).