Calculate the value of a diverter resistance for level compounding in a DC generator, given series field MMF, turns, armature current, and series field resistance.

Step 5: Calculate the value of a diverter resistance (1.4.2)
To achieve level compounding, the series field must provide an MMF of $3600 \text{ AT/pole}$ (calculated in Step 4).
The actual series field coil has $N_{se,actual} = 18$ turns per pole.
The current required to flow through the series field winding ($I_{se}$) to produce this MMF is:

$$I_{se} = \frac{\text{Required MMF}}{N_{se,actual}}$$ $$I_{se} = \frac{3600 \text{ AT/pole}}{18 \text{ turns/pole}}$$ $$I_{se} = 200 \text{ A}$$

The total armature current ($I_a$) is $255.4 \text{ A}$ (calculated in Step 4). This current splits between the series field winding and the diverter resistance.
The current that must flow through the diverter ($I_{div}$) is:

$$I_{div} = I_a - I_{se}$$ $$I_{div} = 255.4 \text{ A} - 200 \text{ A}$$ $$I_{div} = 55.4 \text{ A}$$

The total series field resistance ($R_{se}$) is given as $0.2 \Omega$.
The voltage drop across the series field winding ($V_{se}$) is:

$$V_{se} = I_{se} \times R_{se}$$ $$V_{se} = 200 \text{ A} \times 0.2 \Omega$$ $$V_{se} = 40 \text{ V}$$

Since the diverter resistance is in parallel with the series field winding, the voltage across the diverter is also $40 \text{ V}$.
The value of the diverter resistance ($R_{div}$) is:

$$R_{div} = \frac{V_{se}}{I_{div}}$$ $$R_{div} = \frac{40 \text{ V}}{55.4 \text{ A}}$$ $$R_{div} \approx 0.722 \Omega$$

The value of the diverter resistance is $\text{0.722 } \Omega$.