Diverter resistance for flat compound generator

For flat compounding, the series field must supply the extra field ampere-turns that a shunt generator needs to keep the voltage constant from no load to full load. A diverter in parallel with the series field passes the surplus armature current.

        Ia      +----[Rd]----+
   ---->--------+            +----+---> to load
                +--[ Rse ]---+    |
                   Ise            Rsh (across load)

Step 1: Extra ampere-turns needed

$$AT_{extra} = \Delta I_{sh}\times N_{sh} = 0.2 \times 1000 = 200\ \text{AT per pole}$$

Step 2: Series field current required

$$I_{se} = \frac{AT_{extra}}{N_{se}} = \frac{200}{4} = 50\ \text{A}$$

Step 3: Diverter current

In a long shunt machine the series field (with its diverter) is in the armature circuit, so the full-load armature current of 80 A divides between the series field and the diverter:

$$I_d = I_a - I_{se} = 80 - 50 = 30\ \text{A}$$

Step 4: Diverter resistance

The diverter and series field are in parallel, so their voltage drops are equal:

$$I_dR_d = I_{se}R_{se} \;\Rightarrow\; R_d = \frac{I_{se}R_{se}}{I_d} = \frac{50\times 0.05}{30} = 0.0833\ \Omega$$

Answer: diverter resistance $R_d = 0.0833\ \Omega$.