You can always use a higher voltage rated capacitor, it will probably just last a little longer.
Q = CV, Q is charge, C is capacitance, V is voltage. C= Q/V = dQ/dV since it is linear function = 0.41F
Not all capacitors can be used for 220V; There are capacitors that are specially designed to withstand such high voltages.AnswerThe voltage rating of a capacitor is normally expressed as a d.c. value. If you want to use it on a 220-V a.c. system, then you must take into account that 220 V is an rms value, so you must determine its peak value. The peak value of 220 V (rms) is 311 V. So your capacitor must have a rated value in excess of 311 V d.c., or its insulation will fail.
Use of rvt in capacitor bank
Depending on the circuit, 63% of the available voltage.
In general, no. You need to use the correct capacitor as designed for the circuit.
ceramic capacitor does not simultaneously discharges instead it maintains stead flow of charges through it
You can always use a higher voltage rated capacitor, it will probably just last a little longer.
No
No. The purpose of the zener diode is to clip (turn on) at a certain voltage. A capacitor will not exhibit this behavior.
If the motor is rated at 370 v and the supply is 370 v, the capacitor needs to be rated at 370 or more volts, so a 440 v capacitor will be OK.
Q = CV, Q is charge, C is capacitance, V is voltage. C= Q/V = dQ/dV since it is linear function = 0.41F
Not all capacitors can be used for 220V; There are capacitors that are specially designed to withstand such high voltages.AnswerThe voltage rating of a capacitor is normally expressed as a d.c. value. If you want to use it on a 220-V a.c. system, then you must take into account that 220 V is an rms value, so you must determine its peak value. The peak value of 220 V (rms) is 311 V. So your capacitor must have a rated value in excess of 311 V d.c., or its insulation will fail.
Use of rvt in capacitor bank
For this problem, we use the equation: C=Q/V; where C=capacitor, Q=charge, V=volts We are trying to find Q; the C and V are given, so we just need to plug the numbers into the equation. C=7.90F V=24.0 V Q=? 7.90=Q/24.0 solve for Q
Depending on the circuit, 63% of the available voltage.
No