EMF of short shunt compound generator (6 kW, 250 V)

In a short-shunt compound generator the shunt field is connected directly across the armature, and the series field carries the load current.

Given: $P = 6$ kW, $V = 250$ V, $R_a = 0.1\ \Omega$, $R_{se} = 0.2\ \Omega$, $R_{sh} = 250\ \Omega$.

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

Step 1: Load current

$$I_L = \frac{P}{V} = \frac{6000}{250} = 24\ \text{A}$$

Step 2: Voltage across the shunt field

The series field carries $I_L$, so

$$\begin{aligned} V_{sh} &= V + I_L R_{se} \\ &= 250 + 24 \times 0.2 = 254.8\ \text{V} \end{aligned}$$

Step 3: Shunt field current and armature current

$$\begin{aligned} I_{sh} &= \frac{V_{sh}}{R_{sh}} = \frac{254.8}{250} = 1.0192\ \text{A} \\ I_a &= I_L + I_{sh} = 24 + 1.0192 = 25.0192\ \text{A} \end{aligned}$$

Step 4: Generated emf

$$\begin{aligned} E_g &= V_{sh} + I_a R_a \\ &= 254.8 + 25.0192 \times 0.1 \\ &= 254.8 + 2.502 = 257.30\ \text{V} \end{aligned}$$

Brush drop is not given, so it is neglected.

Answer: EMF generated by the armature $E_g \approx 257.3$ V.