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100 WVDC means 100 working volts DC. That is the maximum operating voltage that the capacitor is certified to have across it. Exceeding that rating could puncture a hole in the dielectric, leading to catastrophic failure of the capacitor.

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How you find capacitance of ceiling fan through formula though it is running 220V 50Hz and as written on capacitor of 2.5uf but how from formulation?

To find the capacitance of a ceiling fan, you can use the formula for capacitive reactance, which is ( X_C = \frac{1}{2\pi f C} ), where ( X_C ) is the capacitive reactance, ( f ) is the frequency in hertz, and ( C ) is the capacitance in farads. Given that the capacitor is rated at 2.5 µF, you can convert this to farads (2.5 µF = 2.5 x 10^-6 F). Using the frequency of 50 Hz, you can calculate the capacitive reactance and confirm that the capacitor value is appropriate for the ceiling fan's operation. However, the primary value of capacitance is already provided by the capacitor itself.


Why capacitor make 90 degree phase shift between voltage and current?

To answer this we assume that the current in a passive component can be written as: i(t) = I*cos(wt +phi), where I is the constant current amplitude for a resistor: V=IR, v(t) = A*R*cos(wt+phi) thus, V = I*R angle(phi) for a capacitor: i(t) = C*(dv/dt) v(t) = V*cos(wt +phi) dv/dt = V*w*sin(wt +phi) therefore: i(t) = wCV*sin(wt + phi) v(t) = V*cos(wt +phi) from this it is clear that the current in a capacitor is 90degrees out of phase (sin->cos 90degrees difference) and that the current amplitude is dependent on capacitance value and frequency (w=2*pi*f). remember because capacitors are not ideal the 90degree phase shift will vary and be dependent on paracitic elements such as parallel/series resistance and series inductance.


Why capacitor short circuit in ac and open circuit in dc analysis?

Reactance of capacitor is inversely proportional to frequency. I should not need to write the exact equation here, its in your textbook. All you need is that its inversely proportional to frequency for proof.We will now assume an ideal capacitor to keep the analysis simple.at DC the frequency is zero, the inverse of this is infinite reactance: open circuitat low frequency AC frequency is low, the inverse of this is high reactanceat midrange frequency AC frequency is midrange, the inverse of this is midrange reactanceat high frequency AC frequency is high, the inverse of this is low reactanceat infinite frequency AC frequency is infinite, the inverse of this is zero reactance: short circuitThis disproves your original statement as written, except for the special cases of DC and infinite frequency AC (which does not occur), for ideal capacitors.As all real capacitors are nonideal, they have leakage resistance. This means that even for the special case of DC the capacitor is not a true open circuit, just a very high resistance resistor. Which also disproves it for the remaining case of DC in real capacitors.


What effect capacitor has on alternating current ac?

In the basic configuration, a capacitor is constructed with two parallel conductor plates with a layer of insulating material in between. When the cap is hooked up to the AC power supply, the voltage (v) across the plates and the charge (q) induced on the plates follow this capacitance expression: C = dq/dv or i = C dv/dt, where C is determined by the properties of the insulating material and the geometry of the cap (in the case of the parallel plates, the separation between the two electrodes (t). For the parallel plates, C can be written as (dielectric constant * plate area / t). Electrically, the change in the charge induced on the plates (dq), is directly related to the change in voltage difference (dv) between the two plates, since C is a constant. Theoretically, no energy is lost by charging and discharging the cap with an AC current. When the cap absorbs electrical energy from the power supply, it stores the energy in the electric field in the insulator. When discharging, the cap gives the stored energy back to the circuit -- hence, no energy loss. In a circuit, we use the cap to prolong/smoothen/resist any voltage change in time or to absorb a sudden energy surge (electrostatic discharge and power-line glitches, for example).


How do you find the voltage on a capacitor?

Charge the capacitor with a voltage source through a resistor. Keep track of how long it takes the capacitor to charge to the voltage level. The value of the capacitor is c=time/5R where R is the value of the resistor and time is the charging time.

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