Power to 10 Ω load from shunt generator

First find the generated emf from the emf equation, then solve the circuit (armature resistance in series with the load and shunt field in parallel) to get the terminal voltage.

Given: $P = 4$, wave winding so $A = 2$, $Z = 378$, $\phi = 0.02$ Wb, $N = 1000$ rpm, $R_a = 1\ \Omega$, $R_{sh} = 100\ \Omega$, $R_L = 10\ \Omega$.

Step 1: Generated emf

$$\begin{aligned} E_g &= \frac{P \phi Z N}{60 A} \\ &= \frac{4 \times 0.02 \times 378 \times 1000}{60 \times 2} = 252\ \text{V} \end{aligned}$$

Step 2: Equivalent resistance across the terminals

Load and shunt field are in parallel:

$$R_p = \frac{R_L R_{sh}}{R_L + R_{sh}} = \frac{10 \times 100}{110} = 9.091\ \Omega$$

   +---Ra=1---+-------+
   |          |       |
  (252 V)   Rsh=100  RL=10
   |          |       |
   +----------+-------+

Step 3: Armature current and terminal voltage

$$\begin{aligned} I_a &= \frac{E_g}{R_a + R_p} = \frac{252}{1 + 9.091} = 24.97\ \text{A} \\ V &= I_a R_p = 24.97 \times 9.091 = 227.03\ \text{V} \end{aligned}$$

Check: $V = E_g - I_a R_a = 252 - 24.97 = 227.03$ V.

Step 4: Power absorbed by load

$$\begin{aligned} I_L &= \frac{227.03}{10} = 22.70\ \text{A} \\ P_L &= \frac{V^2}{R_L} = \frac{227.03^2}{10} = 5154\ \text{W} \end{aligned}$$

Answer: Power absorbed by the load $\approx 5.15$ kW (5154 W), with $E_g = 252$ V and $V = 227.03$ V.