I am having some difficulty in this question:

An inductor operates at a frequency of 50Hz. Inductive Reactance is 250 ohms. I need to calculate the Inductance.

I know the calculation is XL = 2x3.14xFxL to get the inductive resistance. But how would I find the inductance?

 

What does the 'L' represent in that formula?

 

You need to get the L term in that equation all by itself. Do you recall how to do that?

 

Well, then you have:

250 ohm = 2 x 3.14 x 50 Hz x L

And you need to get L. It looks like a very easy thing to do, right?

 

You need to get the L term in that equation all by itself. Do you recall how to do that?

I dont recall how to do it, no.

 

For the formula

Where:
XL = inductive reactance on ohms, Ω
π = Greek letter Pi, 3.14
f = frequency in Hz
L = inductance in henries

 

Let's say I have an equation that tells me that to find my car's mileage I take the distance traveled and divide it by the fuel used, meaning that my equation is

mileage = distance / fuel

Now, let's say that I know that my car's mileage is 20 miles per gallon and I just traveled 400 miles. Can you rearrange that formula to get the fuel used in terms of the mileage and the distance traveled?

Remember in that other thread when we talked about needing to improve algebra skills? This is why. You are not able to focus on learning the electrical engineering concepts because you are being hampered by poor basic math skills. You really need to make a concerted effort to correct that situation, or else you will always struggle to just get a very weak understanding of the engineering because your progress will always be slowed to a crawl by the weak math skills.

This is a quite vicious cycle because, at the same time you are struggling, many (hopefully) of your peers are not (to this degree), which means they will learn the concepts better and be better prepared to take the next course, while you will be increasingly less prepared than them. At some point, the wheels come off and you get left in the dust. Don't let that happen -- get your math skills up to speed.

 

I know the calculation is XL = 2x3.14xFxL to get the inductive resistance.

That's reactance, not resistance.
An inductor can also have an intrinsic resistance, but that's not part of this problem.

Note that when you have an equation like ωL=X (note that your equation is incorrect) you can divide both sides of the equation by the same value to isolate the variable of interest without affecting the correctness of the equation.
Thus to isolate L in the above equation you divide both sides by ω.
This gives (ωL/ω) =(X/ω) or L = (X/ω).

 

I could be wrong but I have had a go:

250ohms divided by 50Hz divided by 6.24 = 0.801A

 

250ohms divided by 50Hz divided by 6.24 = 0.801A

A?

 

The units of inductance are Henrys, not Amperes.

 

I could be wrong but I have had a go:

250ohms divided by 50Hz divided by 6.24 = 0.801A

Remember that little suggestion about tracking units....?

Although, in all fairness, it looks like you are heading in that direction.

1 Hz is 1 cycle/s.

So when you divide 250 Ω by 50 cycles/s, you get 5 Ω·s/cycle. There are 2π radians/cycle. Not that 2π is not 6.24, but much closer to 6.28. Perhaps just a simple typo.

When we divide by this we get L = 0.796 Ω·s

Since I'm assuming you don't have any calculus (I could be wrong), I don't know what the best equation is to use to figure out the relationship between henries and ohm-seconds, so I'll just tell you.

1 H = 1 Ω·s

So multiply

L = (0.796 Ω·s)·(1 Ω·s / 1 H) = 0.796 H = 796 mH

Track your units THROUGH your work, and you won't make the mistake of tacking on the wrong units to an otherwise correct result, because you won't be able to get the units to come out to be amps.

 

L = (0.796 Ω·s)·( 1 H/1 Ω·s ) = 0.796 H = 796 mH

 

This Homework Help. You are not helping anybody giving away the solution.

 

You need to get the L term in that equation all by itself. Do you recall how to do that?

I dont recall how to do it, no.

If you expect to do much of anything in electronics-- anything more than blindly dabbling, that is-- you need to brush up on your algebra. Manipulating simple equations to isolate a single variable on one side needs to be a skill in your everyday mental toolkit; without that, you're completely helpless.

 

If you expect to do much of anything in electronics-- anything more than blindly dabbling, that is-- you need to brush up on your algebra. Manipulating simple equations to isolate a single variable on one side needs to be a skill in your everyday mental toolkit; without that, you're completely helpless.

Thanks for that I will bare that in mind. But then anything is only simple if you know how to do it. I am brushing up on my maths as it’s been rather a long time since I done any which involved algebra.

 

Thanks for that I will bare that in mind. But then anything is only simple if you know how to do it. I am brushing up on my maths as it’s been rather a long time since I done any which involved algebra.

Sounds like a plan. Stick to it, and practice until it becomes second nature.

 

Ok, I have another question.

A coil has an inductance of 0.2H its connected across a 230V, 50Hz supply. It is asking for the current flowing in the circuit if we assume there is no resistance.

I wrote the formula 2*3.14*F*L which was 2*3.14*50Hz*0.2H = 62.8 which then gives me the Inductive Reactance. It says that there is no resistance. So then if I follow OHMS Law I = 230v divided by 62.8 = 3.66A

 

Your answer looks good. And reactance (XL) is measured in the unit of ohms (Ω).

 

I wrote the formula 2x3.14xFXL which was 2x3.14x50Hzx0.2H = 62.8 which then gives me the Inductive Reactance. It says that there is no resistance. So then if I follow OHMS Law I = 230v divided by 62.8 = 3.66A

A suggestion: when writing out a formula or an equation, use an asterisk ("*") to indicate multiplication rather than an "x". It will reduce confusion-- and promote clearer thinking.