A dc series generator has the field resistance of 002 ohms a

A dc series generator has the field resistance of 0.02 ohms and a series field diverter of maximum 0.1 ohm. The armature resistance is 0.04 ohm. If the connected load is a 20 kW load at 200 V, and the induced emf is 205 V, what is the value of diverter resistance?

Solution

Diverter resistance means the resistance in parallel with field to bypass the load current to prevent field and also provide required flux. So diverter resistance and field resistance are in parallel and this combination is in series with armature resistance.

so voltage drop appears across this resistance combination. Total resistance = Ra+(Rf // Rd).

voltage drop = 205-200 = 5V.

Load current = power/voltage = 20000/200 = 100A.

(Ra+(Rf*Rd)/(Rf+Rd))*100 = 5

(0.04+(0.02*Rd)/(0.02+Rd) = 0.05

(0.02*Rd)/(0.02+Rd) = 0.01

0.02*Rd= 0.01*(0.02+Rd)

0.01Rd= 0.0002

Rd= 0.01 ohm.