How to Solve Capacitive Reactance Formula Calculator

This capacitive reactance formula calculator can calculate capacitance from reactance and can calculate the capacitive reactance. Capacitors have an effective resistance known as reactance that varies with frequency. When a direct current (DC) is applied to a capacitor, it acts as an open circuit. But as you apply higher frequencies to a capacitor, its reactance begins to approach a short circuit. It’s important to note that while reactance uses the units of ohms, it’s not the same as resistance. However, you can still perform calculations on it using ohms law. With a purely resistive load, the voltage over the resistor stays in phase with the current going through it, but in a capacitor, the current leads the voltage by 90 degrees for any given frequency. You may also be interested in the inductive reactance calculator.

The calculator below finds the capacitive reactance (X_C) for a given frequency in hertz (Hz).

Capacitive Reactance Formula

The formula used to perform this calculation is below. Remember to convert all units to Farads, Hertz, and Ohms when using the Capacitive Reactance Formula.

Capacitive Reactance Formula:

 where:
     • X_C = Capacitive Reactance (ohms)
     • f = Frequency (hertz)
     • C = Capacitance (farads)

Solution Examples:

We will use the formula above to work out a few real world example problems. You can always check your work with the calculator.

Example #1:

For this example, we will have a 1kHz signal, and a 10uF capacitor. We will be solving for the capacitive reactance of the capacitor.

The first thing that needs to be done is to get all of the values into the correct units. The frequency of 1kHz is 1000Hz, and 10uF is 0.00001 Farads. We will enter the values into the formula as shown.

$\fn_jvn \small X_{C}=\frac{1}{2\pi fC}=\frac{1}{2\pi 1000\cdot 0.00001}=\frac{1}{2\pi 0.01}=15.9\Omega$

The result is that when a signal of 1kHz is applied to this capacitor, it has a reactance of 15.9 ohms.

Example #2:

For this example, we have a 1uF capacitor, and we want to know what frequency we need to get a capacitive reactance of 470 ohms.

As always, the first step is to convert everything to the correct units. The 1uF capacitor is 0.000001 Farads, and the 470 ohms is already in ohms. The formula above solves for capacitive reactance (Xc), so we will need to use algebra to rearrange the variables to solve for the frequency.

$\fn_jvn \small X_{C}=\frac{1}{2\pi fC} \Rightarrow 2\pi X_{C}C=\frac{1}{f} \Rightarrow \frac{1}{2\pi X_{C}C}=f$

So now that we have solved for frequency, we can plug in the variables and solve for f.

$\fn_jvn \small f=\frac{1}{2\pi X_{C}C}=\frac{1}{2\pi 470\cdot 0.000001}=\frac{1}{2\pi 0.00047}\approx 338.6Hz$

The result is that if you want a 1uF capacitor to have a capacitive reactance of 470 ohms, you need to expose it to a frequency of 338.6 Hz.

Example #3:

For this example, he have a signal of 1MHz, and we want to figure out what value of capacitor we need to get a capacitive reactance of 1kohm.

We first need to convert everything to the appropriate units. 1MHz is 1000000 Hz, and 1kohm is 100 ohms. We need to use algebra to convert the formula for capacitive reactance to solve for capacitance.

$\fn_jvn \small X_{C}=\frac{1}{2\pi fC}\Rightarrow 2\pi fX_{C}=\frac{1}{C}\Rightarrow \frac{1}{2\pi fX_{C}}=C$

So now that we have solved for capacitance, we can plug in the variables and solve for C.

$\fn_jvn \small C=\frac{1}{2\pi fX_{C}}=\frac{1}{2\pi 1000000\cdot 100}\approx 0.00000000159 F$

This result is really small. So we need to convert it to appropriate units. 0.00000000159 Farads is equal to 1.59 nF.

The result is that if you have a 1MHz signal, and want a capacitive reactance of 100 ohms, you need to use a capacitor with a value of about 1.59 nF.

Capacitive Reactance Formula

Solution Examples:

Example #1:

Example #2:

Example #3:

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