How to calculate Inductive Impedance???

WBahn

Joined Mar 31, 2012
30,088
Thanks again WBahn, your points are valid, I'm getting there.
No problemo.

It takes effort and lots and lots of practice. One thing to do is forget that calculators exist for the majority of any given problem and work the math manually as much as possible. Use the calculator for sin() and exp() and the like and use it to multiply anything more than two three digit numbers together. Calculators are both our best friend and our worst enemy, so be sure to keep them in check and don't become too dependent on them.

To help develop your estimating ability, get in the habit of estimating the answer (including intermediate answers along the way) of every problem you work. Whenever your answer is more than a factor of five off the final answer, make a point of working with the problem until you understand why your estimate was off -- that will help you learn how to make better estimates in the future.

Just remember, always track your units and always ask if the answer makes sense.
 

vk6zgo

Joined Jul 21, 2012
677
Thanks The Electrician, if I square root 7325494560000 I get 2706565.085.

So that will be 2.7M That looks about right ?? I hope.

Thanks JDR04
I think your problem is that as soon as you write out the formula,you forget that you are trying to analyse an Electrical circuit problem.
Back in High School,you were probably given problems from fields you knew nothing about.

Here,you already know a lot!

(1) It is a Series circuit,so the same current flows through all the components.----This means that the 2.7MΩ is always in circuit,so what does that imply about the minimum value of Z?

(2) The Series Impedance Formula must work for both extremes of XL,
XL=∞,& XL = zero Ω.
Make XL=zero Ω,& substitute that in the formula--what is the value of Z now?

By the way,Z is a bit larger than 2.7MΩ.
It has to be,as XL is not zero Ω.
You also rounded off the value of XL a bit low in your original calculations.
 
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