Given :
Capacitor , C = 55 μF .
Energy is given by :
.
To Find :
The current through the capacitor.
Solution :
Energy in capacitor is given by :
Now , current i is given by :
![i=C\dfrac{dv}{dt}\\\\i=C\dfrac{d[603.02cos(337t)]}{dt}\\\\i=-55\times 10^{-6}\times 603.03\times 337\times sin(337t)\\\\i=-11.18\ sin(337t)](https://tex.z-dn.net/?f=i%3DC%5Cdfrac%7Bdv%7D%7Bdt%7D%5C%5C%5C%5Ci%3DC%5Cdfrac%7Bd%5B603.02cos%28337t%29%5D%7D%7Bdt%7D%5C%5C%5C%5Ci%3D-55%5Ctimes%2010%5E%7B-6%7D%5Ctimes%20603.03%5Ctimes%20337%5Ctimes%20sin%28337t%29%5C%5C%5C%5Ci%3D-11.18%5C%20sin%28337t%29)
( differentiation of cos x is - sin x )
Therefore , the current through the capacitor is -11.18 sin ( 377t).
Hence , this is the required solution .
Answer:
ere el merjor 5iyer
Explanation:
yyhh espero ayuder 8 mucho 666
Answer is: 12.8 because when you multiply it by the 2nd power of 8 then the scraper equals to be 12.8 in height and that’s how much each scraper requires to operate :)
Answer:
The maximum possible volume flow of gasoline is 0.543 m^3/s
Explanation:
Power = pressure differential × volume flow rate
Power = 3.8 kW
Pressure differential = 7 kPa
Volume flow rate = power ÷ pressure differential = 3.8 ÷ 7 = 0.543 m^3/s
