DeepThinker, You were correct. It was me that was wrong.

Les.

NP, we all make mistakes. I made a mistake when I first posted my response and had to go back and correct it myself. I dropped the 20e3 and made it 20 .

 

Hie @LesJones
let me know if I made mistake here,
Current I got is 0.00718A from resistor and Voltage.
From here impedance Z=V/I, 1V/0.00718=139.27
to get value of XL, I used (Z)square= (R)square+ (XL)square
XL, I got is 96.93
using Xl=2*pie*f*L
I got L= 0.000771 =771uH

is this correct?

You got the correct answer but I would recommend while showing your work to be clear on Z verses |Z|, I vs |I| and V vs |V|.
|x| is the absolute magnitude of the complex number (that is the hypotenuse on a graph)
x = is a complex number.

Example:
|Z|=sqrt(R^2+XL^2)
and
Z=R+jXL

 

this may help--

Inductance Calculations
Author: Frederick W. Grover
ISBN-10: 048647440-2

 

this may help--

Inductance Calculations
Author: Frederick W. Grover
ISBN-10: 048647440-2

But isnt that a book about calculating inductance from physical constructs?
He needs to understand inductance from a circuit analysis point of view.
In other words, not the inductance knowing that the inductor is 2 inches long and 1 inch wide and has 100 turns, but rather what the voltage across the resistor is when the inductor is 10 Henries and the resistor is 10 Ohms and the applied voltage is 25 volts at 1kHz.
Those are two very very different questions and so the approaches are very very different where one involves physical length measurements and the other electrical measurements.

 

I am not that familiar with the AC analysis, that's why I asked the question here to get something. I have some understanding about impedance and reactance, and my doubt here is, since in this circuit there is no XL or Z value is given then how will I get the value of L? Can you Please tell the steps to find out ? The question here in what value of inductor we need to use in this circuit to make the input voltage 1V(p-p) to 0.718V(p-p) in the output. Thanks

Hello again,

I got a value of (i wont specify the exact value just yet) of 2mH or 3mH or close to one of those.

Did you remember that the total reactance is:
X=sqrt(R^2+XL^2)

It looks like you may have forgotten the square root.

But there are a few different ways to do this. I guess you are most familiar with the reactance method but we could look at other methods too. Do you know how to use complex numbers? That would make this even simpler.

Here is the drawing cleaned up a bit more and reduced in size to take up less room. Notice the very small file size compared to the original.

 

In the old days before pocket calculators, when engineers carried a slide rule & drafting tools, we would solve this with a phasor diagram drawn on radial graph paper (a good drawing could achieve slide rule accuracy).

For this problem, use phasor voltages. Draw a circle with the radius representing the supply voltage and another circle with radius representing the resistor voltage. Draw a line at the 0° mark to be the supply voltage phasor (RED). Now draw a line from the tip of the RED phasor tangent to the BLUE circle. Use a right triangle to draw a perpendicular from the tangent line to the center (BLUE). Measure the length of the GREEN line.

The length of the blue line is proportional to the resistance; the length of the green line is proportional to the reactance of the coil. Use your slide rule to find the inductance of the coil.

 

#1 you didn't say if the circuit is AC or DC. given that you don't know how to analyze AC, I would say ... continue reading your book.

#2 you didn't say what your objective is. using an inductor to "step down voltage" from 1V to 718mV isn't ever prescribed

#3 your using inductor calculation after saying you don't know about AC analysis

ok but your formula will be wrong if your GROUND isn't perfect.

Do you have a perfect ground. Did you study what GROUND is yet? they tell you it's a symbol and it's ground ... but what IS IT? what is it if it's DC? what IS IT if it's AC? Some DC are actually grounded and some not and both use the ground symbol btw.

When you begin specifying "wave forms" you'll find out why a chinese wave form generator (just $35 ?) has allot of circuitry.

You can't get a (good) waveform out of a few parts. It's not going to be that easy. If you see some "buck converter" I mean sure that works: but the factory made ones are efficient do to tight design. Just throwing together some parts likely won't get you a good waveform. It's not that easy.

"inductive loads tend to cause problems with waveforms on a bus" - whatever that means. it means it's not how you step down voltage. I'm sure of that.