How is capacitor connected?

Answer 1

A capacitor joined with a sinusoidal voltage source will bring about a relocation current to move through it. For the situation that the voltage source is V0cos(ωt), the displacement current can be communicated as:

I = C \frac{dV}{dt} = - \omega {C}{V_\text{0}}\sin(\omega t)

At sin(ωt) = - 1, the capacitor has a most extreme (or top) current whereby I0 = ωCV0. The proportion of crest voltage to crest current is because of capacitive reactance (signified XC).

X_C = \frac{V_\text{0}}{I_\text{0}} = \frac{V_\text{0}}{\omega C V_\text{0}} = \frac{1}{\omega C}

XC methodologies zero as ω methodologies endlessness. In the event that XC approaches 0, the capacitor looks like a short wire that emphatically passes current at high frequencies. XC approaches boundlessness as ω methodologies zero. On the off chance that XC approaches unendingness, the capacitor looks like an open circuit that ineffectively passes low frequencies.

The current of the capacitor may be communicated as cosines to better contrast and the voltage of the source:

I = - {I_\text{0}}{\sin({\omega t}}) = {I_\text{0}}{\cos({\omega t} + {90^\circ})}

In this circumstance, the current is out of stage with the voltage by +π/2 radians or +90 degrees.

Answer 2

A; The only benefit is that twice the voltage rating of a single capacitor can handle. the disadvantage is the total capacitance is reduced by one half if equal. to insure proper operation a resistor must be added to each to stabilize the leakage current

Answer 3

I suspect you mean "...when capacitors are connected in series?"

If so, the total capacitance decreases. Calculate as follows:

Ctot= 1/(1/C1+1/C2+1/C3...etc)

Answer 4

Because capacitance of capacitor increases when we connect capacitor parallel to each other & Series connection of capacitor use to reduce the capacitance.