# How do I find C & L from XC and XL?

Calculating Capacitance and Inductance from Capacitive Reactance and Inductive Reactance

I'm trying to find the inductor (L) and capacitor (C) values using the inductive reactance (XL) and capacitive reactance (XC). Edit: I also have the frequency (f=50 Hz).

I know $XL = 2\pi f L$ and I know $XC = \dfrac{1}{2\pi f C}$, but I am not the best at reversing formulas, I am still learning.

Can anyone confirm the correct formulas please?

I have tried assigning random values to each part of the equations, but my answers don't seem to match up for some reason. I don't know where I am going wrong, but nevertheless keen to learn and willing to ask.

I have also tried searching online, but the only articles I can find of relevance are about calculating XC and XL, not C and L from XC and XL.

• Do you have the frequency? Commented Oct 21, 2016 at 20:43
• This question might be better suited for a basic algebra site rather than EE. Commented Oct 21, 2016 at 20:48
• Then I afraid you will have to go back to a very basic math books/classes in order to proceed with electrical engineering.. Commented Oct 21, 2016 at 20:48
• I am not mean, you really need it, take it as a friendly advice. You can't do any engineering without knowing how to solve a basic linear equation with a single unknown. Commented Oct 21, 2016 at 20:52
• Sometimes, we all stand on our own power line. But no kidding, math is getting much much worse from here. But there's hope if you are allowed to use a self-written cheat-sheet in exams. Commented Oct 21, 2016 at 22:40

Where

• $f$ is frequency in hertz
• $C$ is capacitance in farads
• $X_C$ is capacitive reactance in ohms
• $L$ is inductance in henrys, and
• $X_L$ is inductive reactance in ohms,

$$X_L = 2 \ \pi \ f\ L,$$

then divide both sides of the equation by $2\pi\ f$ in order to isolate the $L$, you'll get:

$$\frac {X_L}{2 \ \pi \ f}= \frac{ \require{cancel} \cancel { 2 \ \pi \ f}\ L}{ \require{cancel} \cancel { 2 \ \pi \ f}}\text{ .}$$

Move the $L$ over to the left hand side and, like magic, Voila! $$L= \frac{X_L}{2 \ \pi \ f}\$$

$$Xc = \frac{1}{2 \ \pi \ f\ C}$$
$$C\ X_C = \frac{1\require{cancel} \cancel {C}}{2 \ \pi \ f\require{cancel} \cancel {C}}$$
Finally, in order to isolate C and put the reactive term in its place, divide through by$X_C$ and you'll get:
$$\frac{C \require{cancel} \cancel {X_C}}{\require{cancel} \cancel {X_C}} = \frac{1}{2\pi\ f\ X_C}{ \text{ , }}$$ and after you clean it all up you'll have:
$$C = \frac{1}{2\pi\ f\ X_C}$$