Wave-wound shunt generator, power absorbed by load

The load (10 Ω) and the field winding (100 Ω) are both across the armature terminals, so they are in parallel. The generated emf comes from the emf equation for a wave winding ($A = 2$).

Step 1: Generated emf

$$E = \frac{\Phi Z N P}{60A} = \frac{0.02\times 378\times 1000\times 4}{60\times 2} = 252\ \text{V}$$

Step 2: Terminal resistance and armature current

Load in parallel with field:

$$R_p = \frac{100\times 10}{100 + 10} = 9.091\ \Omega$$ $$I_a = \frac{E}{R_a + R_p} = \frac{252}{1 + 9.091} = 24.973\ \text{A}$$

Step 3: Terminal voltage

$$V = I_aR_p = (24.973)(9.091) = 227.03\ \text{V}$$

Step 4: Load current and power

$$I_L = \frac{V}{10} = 22.703\ \text{A}, \qquad I_f = \frac{V}{100} = 2.270\ \text{A}$$

(Check: $I_L + I_f = 24.973\ \text{A} = I_a$.)

$$P_{load} = \frac{V^2}{R_L} = \frac{(227.03)^2}{10} = 5154\ \text{W}$$

Answer: power absorbed by the load $\approx 5154\ \text{W}$ ($\approx 5.15$ kW) at a terminal voltage of $227.03\ \text{V}$.