# Capacitive reactance formula derivation and calculation

**Capacitive reactance equation derivation**

Capacitive reactance (Xc) is a measure of the opposition to current flow in a capacitive circuit. It is caused by the electric field that is generated between the plates of a capacitor when a voltage is applied across it. The mathematical expression for capacitive reactance is given by the following equation:

Xc = 1 / (2πfC)

Where: Xc = capacitive reactance (Ω) f = frequency of the current (Hz) C = capacitance of the circuit (Farads) π = pi (approximately 3.14)

The derivation of this equation is based on the relationship between the voltage and the current in a capacitive circuit.

In a capacitive circuit, the voltage and current are out of phase by 90 degrees. The current leads the voltage in a capacitive circuit. The relationship between the voltage and the current can be expressed mathematically as:

i = C * dv/dt

Where: i = current flowing through the circuit (Amperes) C = capacitance of the circuit (Farads) dv/dt = rate of change of voltage across the circuit (volts/second)

We can apply the above expression in an AC circuit as shown in the below video,

We get the final equation for capacitive reactance, Xc = 1 / (2πfC). Where f is the frequency of the current flowing through the circuit.

The above equation is the final mathematical expression for capacitive reactance, which is a measure of the opposition to current flow in a capacitive circuit.

**Capacitive reactance calculation**

Capacitive reactance calculation for a given frequency and capacitance.

Eg:- Frequency =50hz and capacitance = 1μf.

To calculate the capacitive reactance (Xc) of a circuit with a capacitance (C) of 1 micro farad (μf) and frequency (f) of 50 Hz, we can use the equation:

Xc = 1 / (2πfC)

Plugging in the given values, we have: Xc = 1 / (2π (50) (1 x 10^-6))

The unit of capacitance is Farad (F) but in this case, the value is given in Microfarads (μf) which is 10^-6 Farads.

Now, calculating this we get: Xc = 1 / (2π (50) (1 x 10^-6)) = 1 / (2π (50) (0.000001)) = 1 / (314 x 0.000001) = 1 / 0.0314 = 31.83 ohms

So the capacitive reactance of the circuit is 31.83 ohms.

But how do you make C the subject of formula?

Could you please explain more.