EMF of 25 kW short shunt generator with brush drop

Given: $P = 25$ kW, $V = 250$ V, $R_{sh} = 125\ \Omega$, $R_{se} = 0.5\ \Omega$, $R_a = 0.1\ \Omega$, total brush drop $V_b = 2$ V; short-shunt connection.

     +--Ra--+-------Rse-------+
     |      |                 |
    (Eg)   Rsh              Load 250 V
     |      |                 |
     +------+-----------------+

Step 1: Load current

$$I_L = \frac{25000}{250} = 100\ \text{A}$$

Step 2: Voltage across shunt field

$$V_{sh} = V + I_L R_{se} = 250 + 100 \times 0.5 = 300\ \text{V}$$

Step 3: Shunt field and armature current

$$\begin{aligned} I_{sh} &= \frac{300}{125} = 2.4\ \text{A} \\ I_a &= I_L + I_{sh} = 100 + 2.4 = 102.4\ \text{A} \end{aligned}$$

Step 4: Generated emf

$$\begin{aligned} E_g &= V_{sh} + I_a R_a + V_b \\ &= 300 + 102.4 \times 0.1 + 2 \\ &= 300 + 10.24 + 2 = 312.24\ \text{V} \end{aligned}$$

Answer: EMF generated in the armature $E_g = 312.24$ V.