The correct answer is $\boxed{\text{D}}$.
The terminal voltage of a shunt generator is given by the following equation:
$V_T = E_A – I_A R_A$
where:
- $V_T$ is the terminal voltage
- $E_A$ is the induced emf
- $I_A$ is the armature current
- $R_A$ is the armature resistance
In this case, we are given that $E_A = 600 \text{ V}$, $I_A = 200 \text{ A}$, and $R_A = 0.1 \Omega$. Substituting these values into the equation, we get:
$V_T = 600 \text{ V} – 200 \text{ A} \cdot 0.1 \Omega = 580 \text{ V}$
Therefore, the terminal voltage of the shunt generator is 580 V.
Option A is incorrect because it is the induced emf, not the terminal voltage.
Option B is incorrect because it is the induced emf plus the armature drop. The armature drop is the voltage drop across the armature resistance, and it is given by the following equation:
$I_A R_A = 200 \text{ A} \cdot 0.1 \Omega = 20 \text{ V}$
Therefore, the armature drop is 20 V, and the induced emf plus the armature drop is 600 V + 20 V = 620 V.
Option C is incorrect because it is the induced emf.
Option D is correct because it is the terminal voltage.