Question

A 75.0 kW, 230-V shunt generator has a generated emf of 243.5 V. If the field current is 12.5 A at
rated output, what is the armature resistance? (Ans. 0.0399).

Answer 1

Answer

Shunt current

$I_s=\frac{P}{V}=\frac{75000}{230}=326A$

So $R_s=\frac{V}{I}=\frac{230}{326}=0.7\Omega$

So using formula

$I_s=\frac{V}{R_s+R_A}$

Where

R_A is armature resistance. Putting all value then we get

R_A=0.0399$\Omega$