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:
I would like to help can you take a picture without the bot telling you something. Maybe then I can figure it out.
Answer:
Tension in cable BE= 196.2 N
Reactions A and D both are 73.575 N
Explanation:
The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence
hence

Therefore, tension in the cable, 
Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then



Similarly,


Therefore, both reactions at A and D are 73.575 N
Answer:
8.24μm
Explanation:
The theory of brittle fracture was used to solve this problem.
And if you follow through with the attachment made a the subject of the formula
Such that,
a = 2x(69x10⁹)x0.3/pi(40x10⁶)²
= 4.14x10¹⁰/5.024x10¹⁵
= 8.24x10^-06
= 8.24μm
This is the the maximum length of the surface flaw
search it and you will get on internet