Short shunt vs long shunt generated voltage (150 A, 230 V)

Given: $I_L = 150$ A, $V = 230$ V, $R_a = 0.15\ \Omega$, $R_{se} = 0.1\ \Omega$, $R_{sh} = 100\ \Omega$, brush drop 2 V per brush, so total brush drop $= 2 \times 2 = 4$ V (one positive and one negative brush in the current path).

(a) Short-shunt connection

     +--Ra--+-------Rse-------+
     |      |                 |
    (Eg)   Rsh              Load 230 V
     |      |                 |
     +------+-----------------+

$$\begin{aligned} V_{sh} &= V + I_L R_{se} = 230 + 150 \times 0.1 = 245\ \text{V} \\ I_{sh} &= \frac{245}{100} = 2.45\ \text{A} \\ I_a &= I_L + I_{sh} = 150 + 2.45 = 152.45\ \text{A} \\ E_{g,s} &= V_{sh} + I_a R_a + V_{b} \\ &= 245 + 152.45 \times 0.15 + 4 \\ &= 245 + 22.87 + 4 = 271.87\ \text{V} \end{aligned}$$

(b) Same machine as long shunt (same load: 150 A at 230 V)

     +--Ra--Rse--+------------+
     |           |            |
    (Eg)        Rsh         Load 230 V
     |           |            |
     +-----------+------------+

Now the shunt field is across the terminals and the series field carries $I_a$:

$$\begin{aligned} I_{sh} &= \frac{230}{100} = 2.3\ \text{A} \\ I_a &= 150 + 2.3 = 152.3\ \text{A} \\ E_{g,l} &= V + I_a (R_a + R_{se}) + V_b \\ &= 230 + 152.3 \times 0.25 + 4 \\ &= 230 + 38.075 + 4 = 272.075\ \text{V} \end{aligned}$$

Ratio of generated voltages:

$$\frac{E_{g,l}}{E_{g,s}} = \frac{272.075}{271.8675} = 1.00076$$

Answer: (a) $E_g = 271.87$ V (short shunt). (b) $E_g = 272.08$ V as long shunt; ratio long : short $= 1.00076$ (about 1.0008 : 1). The long-shunt machine needs a slightly higher emf because the series field now carries the larger armature current.