a wind electric generator is generating electrical power which flow an electric current of 1.2 A through a circuit of 50 ohm. if the power of wind is 300 W. then the efficiency of wind mill is:
A) 24%
B)12%
C)35%
D)20%
Answers
Efficiency of a body is the percentage of the work done at input being converted to the work done at output .
Power of the wind mill is 300 W .
We know that the power of the body is the product of potential difference and the current .
P = V I .
By Ohm's Law we have V = I R :
P = ( I R ) × I
⇒ P = I²R
Current flowing ( I ) = 1.2 A .
Resistance of the circuit ( R ) = 50 Ω
P = ( 1.2 )² × 50 W
⇒ P = 1.44 × 50
⇒ P = 72
The power generated or the power output is 72 W .
Let the time for which the work is done be T .
Power is the rate of doing work which is :
P = W / T
⇒ W = P×T
Work input = 300 × T ⇒ 300 T
Work output = 72 × T ⇒ 72 T
Efficiency = work output / work input × 100 %
⇒ η = 72 T / 300 T × 100 %
⇒ η = 72 / 300 × 100 %
⇒ η = 72/3 %
⇒ η = 24 %
OPTION A
The efficiency of windmill is 24 % .
current (i)= 1.2 A
Resistance(R)= 50 ohm
P= i^2 R
= ( 1.2)^2 (50)
= 1.44× 50
= 72 W
This is generated power
power of wind = 300
Now here efficiency asked that is how much efficient it is
So % efficiency = 72/300. × 100
= 72/3 = 24%
Thanks