The e.m.f. induced in the armature of a shunt generator is 600 V. The armature resistance is 0.1 ohm. If the armature current is 200 A, the terminal voltage will be

To find the terminal voltage ((V_t)) of the generator, we can use the equation:

[ V_t = E_g - I_a cdot R_a ]

Where:
- ( E_g ) is the generated electromotive force (EMF) induced in the armature.
- ( I_a ) is the armature current.
- ( R_a ) is the armature resistance.

Given:
- ( E_g = 600 ) V
- ( R_a = 0.1 ) ohm
- ( I_a = 200 ) A

Substituting the given values into the equation:

[ V_t = 600 , text{V} - 200 , text{A} times 0.1 , Omega ]
[ V_t = 600 , text{V} - 20 , text{V} ]
[ V_t = 580 , text{V} ]

Therefore, the terminal voltage will be 580 V.

So, the correct answer is:

D. 580 V