A dc shunt generator delivers 195 A at terminal voltage of 200 V- armature resistance is 0-03 - and field winding resistance is 40 - The stray losses are 900 W- The efficiency of generator is

I_f=20040=5A I_a=I_f+I_L=195+5=200 A Copper losses =I_a^2R_a+I_f^2R_f =(200)^2×0.03+5^2×40=2200 W Output power = V_tI_L=195×200=39000 W Input power = Outpu ...

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Standard XII

Physics

Impedance in RLC Circuit

A dc shunt ge...

Question

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

89.14%

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97.14%

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86.14%

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92.63%

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Solution

The correct option is D 92.63%
$I_{f} = \frac{200}{40} = 5 A$

$I_{a} = I_{f} + I_{L} = 195 + 5 = 200 A$

$Copper losses = I_{a}^{2} R_{a} + I_{f}^{2} R_{f}$

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

$Output power = V_{t} I_{L} = 195 \times 200 = 39000 W$

$Input power = Output power + Cu losses + Stray losses$

$= 39000 + 2200 + 900 = 42100 W$

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

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

The correct option is D 92.63%If=20040=5A Ia=If+IL=195+5=200A Copper losses =I2aRa+I2fRf =(200)2×0.03+52×40=2200 W Output power = VtIL=195×200=39000 W Input power = Output power + Cu losses + Stray losses =39000+2200+900=42100 W Efficiency, η=outputinput×100% =3900042100×100%=92.63%

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