4-pole shunt generator, wave wound, 3 kW

In a shunt generator the armature current supplies both the load and the field: $I_a = I_L + I_{sh}$. The generated emf must also cover the armature drop and the brush drop.

(a) Armature current

Load current:

$$I_L = \frac{P}{V} = \frac{3000}{120} = 25\ \text{A}$$

Shunt field current:

$$I_{sh} = \frac{V}{R_{sh}} = \frac{120}{60} = 2\ \text{A}$$ $$I_a = I_L + I_{sh} = 25 + 2 = 27\ \text{A}$$

(b) Current in each armature conductor

A wave winding has $A = 2$ parallel paths regardless of the number of poles.

$$I_{\text{conductor}} = \frac{I_a}{A} = \frac{27}{2} = 13.5\ \text{A}$$

(c) Generated emf

Brush drop $= 1\ \text{V} \times 2\ \text{brushes (one positive, one negative)} = 2\ \text{V}$.

$$\begin{aligned} E &= V + I_aR_a + V_{brush} \\ &= 120 + (27)(0.2) + 2 \\ &= 120 + 5.40 + 2 = 127.40\ \text{V} \end{aligned}$$

Answer: (a) $I_a = 27\ \text{A}$; (b) current per conductor $= 13.5\ \text{A}$; (c) $E = 127.40\ \text{V}$.