Question

A dc shunt generator delivers 195 A at terminal voltage of 200 V, armature resistance is $0.03\Omega$ and field winding resistance is $40\Omega$. The stray losses are 900 W. The efficiency of generator is

• A. 89.14%
• B. 97.14%
• C. 86.14%
• D. 92.63%

Answer 1

The correct option is D 92.63%

$I_f=\frac{200}{40}=5A$

$I_a=I_f+I_L=195+5=200A$

$\text{Copper losses}=I_a^2{R}_a+I_f^2{R}_f$

$=(200{)}^2\times 0.03+5^2\times 40=2200W$

$\text{Output power =}{V}_t{I}_L=195\times 200=39000W$

$\text{Input power = Output power + Cu losses + Stray losses}$

$=39000+2200+900=42100W$

$\text{Efficiency,}\eta =\frac{\text{output}}{\text{input}}\times 100\%$
$=\frac{39000}{42100}\times 100\%=92.63\%$