**CAPACITANCE AND CAPACITIVE REACTANCE**

BEHAVIOUR OF CAPACITORS

**CAPACITORS**

A capacitor is nothing more than two conductive plates or surfaces separated by a non-conducting dielectric. If voltage is applied to the two plates, one plate will develop a positive charge and the other negative. Like charges attract so this charge will remain so long as the applied voltage remains the same. If the voltage is now changed, the current that flows will be whatever current needs to flow to reach the new voltage.

If the size of the plates is increased or the distance between them is decreased, the electrical attraction of the oppositely charged plates will increase thus increasing the capacitance.

The charge stored in a capacitor represents stored energy which is available for return to the circuit when the voltage changes. A perfect capacitor therefore does not dissipate power because energy stored in one half a cycle is returned to the circuit in the other half. There is, of course, no such thing as a perfect capacitance. There will always be the resistance of the materials making up the capacitor and some leaking of charge either through the dielectric or the air.

**VOLTAGE AND CURRENT**

As stated, a capacitor will charge when a voltage is applied and current will flow if the voltage changes. If a voltage is immediately applied to a perfect capacitor, the current will be infinite because the perfect capacitor has zero resistance. The current will depend on the __rate of change__ in voltage and be in the same direction. That is, when the voltage is decreasing the current will be negative. When the voltage is increasing, the current will be positive. Consider the following diagram:-

NOTE:- The initial current was established in previous cycles. At point A, the voltage is at a maximum, but its rate of change is ZERO. The current will therefore be ZERO. At point B, the rate of change of voltage is at a maximum so the current in the capacitor will be at a maximum in the direction of the voltage change. Since the voltage is decreasing, the current will be maximum negative.

**Rule:- In a capacitive circuit, the applied voltage lags the current. For a perfect capacitor this will be by and angle of ^{Π}/2 (90^{o}). Since the voltage lags the current it is said to lead by -^{Π}/2**

**CAPACITANCE AND CAPACITIVE REACTANCE**

If the size of the capacitor is increased, there will be more surface area for current to flow into and a greater force of attraction between the two surfaces and so there will be an increase in capacity. The need to charge will be represented as an opposition to a change in voltage. Capacity is measured in Farads.

The reactance (impedance to current flow) will be directly proportional to the inverse of capacity. Since a higher frequency means a faster rate of change of voltage, it also means higher currents. As capacity increases, the reactance decreases. The formula for capacitive reactance is given by X_{C} = - 1 / ω x C, where X_{C} is the reactance in ohms, ω is the angular frequency in radians per second (2Πf) and C is the capacity. Capacitive reactance is -ve because the current leads the voltage.

NOTE:- Reactance and impedance are two different things. Reactance only has magnitude (including sign) and can be used to calculate the current through a coil, capacitor or combination provided the phase angle is always ±^{Π}/2. Impedance is a complex quantity including both reactive and resistive components even though one component might be 0. In the case of a purely reactive circuit, Impedance = Reactance. In the case of a purely resistive circuit Impedance = Resistance. In any other case |Impedance| = √ ( reactance^{2} + resistance^{2} ) and the phase angle of resulting impedance is arctan(reactance/resistance).

**CAPACITORS IN SERIES**

Since the capacitors are in series, the current through all of them will be the same. (Kirchhoff's first law.) The voltage across each capacitor will be given by the current multiplied by the reactance.

V_{C1} = I x X_{C1} ... WHILE ... V_{C2} = I x X_{C2} ... GIVING ... V = I x X_{C1} + I x X_{C2}

Where

V is the total voltage,

V_{C1} is the voltage across capacitor 1,

V_{C2} is the voltage across capacitor 2,

I is the common current,

X_{C1} is the reactance of capacitor 1 and

X_{C2} is the reactance of capacitor 2.

(NOTE: Because the current leads the voltage, the values of X_{C} will be negative numbers. This makes no difference to calculations.)

The total reactance, X_{C}, will depend on the total current and total voltage thus:-

X_{C} = V / I

Substituting V for I x X_{L1} + I x X_{L2} we get:-

X_{C} = ( I x X_{C1} + I x X_{C2} ) / I ... OR ... X_{C} = I x ( X_{C1} + X_{C2} )/ I

Finally cancelling out the current, "I", we get the formula for combined reactance of two capacitors in series:-

X_{C} = X_{C1} + X_{C2}

Substituting back for capacitance:-

-1 / (ω x C ) = -1 / ( ω x C_{1} ) + -1 / ( ω x C_{2} ) ... OR ... 1 / C = 1 / C_{1} + 1 / C_{2}

Inverting gives:-

C = 1 / ( 1 / C_{1} + 1 / C_{2} )

**CAPACITORS IN PARALLEL**

Since the capacitors are in Parallel, the voltage across both will be the same. The total current will be the sum of currents in both thus:-

I_{C1} = V / X_{C1} ... WHILE .... I_{C2} = V / X_{C2} ... GIVING ... I = V / _{C1} + V / X_{C2}

Where

I is the total current,

I_{C1} is the current through C1,

I_{C2} is the current through C2,

X_{C1} is the reactance of C_{1},

X_{C2} is the reactance of C_{2} and

V is the applied voltage.

The total reactance will be:-

X_{C} = V / I

Substituting for I gives:-

X_{C} = V / ( V / X_{C1} + V / X_{C2} ) ... __OR__ ... V / V ( X_{C1} + X_{C2} )

And cancelling out the V we get:-

X_{C} = 1 / ( 1 / X_{C1} + 1 / X_{C2} )

The value of parallel capacitance can be worked out from further substitution:-

C = C_{1} + C_{2}

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