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:
b. spark plugs
Explanation:
Diesel engines are characterized in that the mechanism that activates the explosion of fuel is high pressures, so when the piston reaches the top of the cylinder, the pressure of the air-diessel mixture is so high that it causes the explosion, this It is what generates the power in a diesel engine.
Answer:
Permanent felony conviction on your criminal record.
Financial restitution if DUI involves an accident, property damage or personal injury
Up to 50 hours community service or fee of $10 per hour of service.
Minimum fine of upto $1000 or $2000, that is if your blood alcohol level was 0.15% or more than this, or you were driving with a minor in the vehicle.
Imprisonment of upto 9 months, if the blood alcohol level was over 0.15% or idf the DUI resulted in ana accident or crash, jail time is increased to one year.
Mandatory imprisonment for not less than 10 days if your second DUI conviction occurs within 5 years of the date of the prior DUI conviction.
180 days to one year probation.
Maximum of 5 years driver's license revocation.
Up to 30 days impoundment of vehicle.
Ignition interlock device.
Criminal record stating adjudication of guilt.
Explanation:
To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are
Real power in 3 phase
Now the Phase Voltage is,
The current phase would be,
Rearranging,
Replacing,
Therefore the current per phase is 2.26kA
