Compound generator emf, long and short shunt

Brush drop $= 1\ \text{V}\times 2\ \text{brushes} = 2\ \text{V}$. Load current $I_L = 50$ A, terminal voltage $V = 500$ V.

(i) Long shunt

The shunt field is across the load terminals:

$$I_{sh} = \frac{V}{R_{sh}} = \frac{500}{250} = 2\ \text{A}, \qquad I_a = I_{se} = 50 + 2 = 52\ \text{A}$$ $$\begin{aligned} E &= V + I_a(R_a + R_{se}) + V_{brush} \\ &= 500 + 52(0.05 + 0.03) + 2 \\ &= 500 + 4.16 + 2 = 506.16\ \text{V} \end{aligned}$$

(ii) Short shunt

The series field carries only the load current. The shunt field is across the armature, so its voltage is

$$V_{sh} = V + I_LR_{se} = 500 + 50(0.03) = 501.50\ \text{V}$$ $$I_{sh} = \frac{501.50}{250} = 2.006\ \text{A}, \qquad I_a = I_L + I_{sh} = 50 + 2.006 = 52.006\ \text{A}$$ $$\begin{aligned} E &= V_{sh} + I_aR_a + V_{brush} \\ &= 501.50 + (52.006)(0.05) + 2 \\ &= 501.50 + 2.600 + 2 = 506.10\ \text{V} \end{aligned}$$

Answer: (i) long shunt $E = 506.16\ \text{V}$; (ii) short shunt $E = 506.10\ \text{V}$.