Capacitive Reactance Formula

Capacitive Reactance Formula: Solved With an Example

This article aims to explore how capacitive reactance can be calculated by following its formula. To analyze the Capacitive Reactance formula, we will have a minimal definition of capacitive reactance and its effects on a circuit. We will also calculate a problem by the Capacitive Reactance Formula of its implications in wireless devices.

What Is Capacitive Reactance?

A capacitor with a positive charge, Q(C), creates an electric field that opposes a decrease in voltage across the capacitor when a constant current I is flowing through it. This phenomenon is called capacitive reactance and affects the RC time consistent related to resistance and capacitance when alternating current passes through a circuit.

Capasitivie Reactance Xc= 1/2πfC

Capacitive Reactance Formula


Xc = Capacitive Reactance in Ohms, (Ω)
π (pi) = a numeric constant of 3.142
ƒ = Frequency in Hertz, (Hz)
C = Capacitance in Farads, (F)

Capacitive Reactance Circuit Diagram With Calculation

A Capacitive Voltage Divider circuit is a passive linear network consisting of a series capacitor and a resistor.

Capacitive Reactance Circuit Diagram
Capacitive Reactance Circuit


Look at the circuit diagram carefully. This circuit has a 10V AC voltage supply across the two different series-connected capacitors of additional capacitance C1 and C2. The charge of each capacitor Q will be the same because of the series connection, but the voltage across them will be different and related to their capacitance values, as V = Q/C.

So, here in the circuit, C1 is a capacitor of 10uF, and C2 is the capacitor of 22 uF; we will now calculate the RMS voltage drop of each capacitor at 10 volts RMS at 80Hz.

Capacitive Reactance of C1:

Xc1= 1/2πfC

= 1/2π*80*10*10^-6
= 200Ω

Capacitive Reactance of C1:

Xc2= 1/2πfC

= 1/2π*80*22*10^-6
= 90Ω

Total Capacitive Reactance Xc= Xc1+Xc2

= 200Ω+90Ω
= 290Ω

So current supply in the circuit Ic= E/Xc

= 10V/290Ω
= 34.5mA

Now, voltage drop on Vc1= I*Xc1

= 34.5mA*200Ω
= 6.9 V

Voltage drop on Vc2= I*Xc2

= 34.5mA*90Ω
= 3.1 V

So, we learn from the calculation that supply voltage and supply frequency will finally be the same for series connections.

And, the most important thing is that the lower value capacitor charges itself higher than, the higher valued capacitor.


About the Author: BDElectricity Staff

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