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
A. 640 V
B. 620 V
C. 600 V
D. 580 V
Answer: Option D
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Related Questions on D.C. Generators
A cumulatively compounded long shunt generator when operating as a motor would be
A. Cumulatively compounded long shunt
B. Differentially compounded long shunt
C. Cumulatively compounded short shunt
D. Differentially compounded short shunt
A. Demagnetization only
B. Cross magnetization as well as magnetization
C. Cross-magnetization as well as demagnetizing
D. Cross magnetization only
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