Taking your example. To make it from 0.5V to 0.359V , The inductor as a resistance it will be 254.6 ohm. After that how to get the value of the inductor? I mean in inductancehi DT,
Consider what would the resistance offered by the Inductor have to be to reduce 0.5Vpk t0 0.359Vpk.
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Also keep in mind that when a AC voltage does not specify peak to peak it is assumed to be RMS. Thus the output should be marked 718mv (p-p).
Since this is an AC analysis problem, what techniques have you learned in the past for handling AC circuits?
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. ThanksSince this is an AC analysis problem, what techniques have you learned in the past for handling AC circuits?
There are several different ways to do this so it would be good if you mentioned how you did other problems in the past unless you want to learn new techniques.
Hie @LesJonesYou know the value of the resistor and you know the voltage across it so you can calculate the current trough it. So you also know the current through the total impedance (Z) formed by XL and R in series. As you also know the source voltage you can calculate Z. From the formula for resistance in series with reactance you can insert the two values you have (R & Z) and then re arrange the formula to make XL the subject. That will give you XL. You then just need to convert XL into L.
Les.
Hie @LesJones
let me know if I made mistake here,
Current I got is 0.00718A from resistor and Voltage.
I agree.
From here impedance Z=V/I, 1V/0.00718=139.27 ohms
I agree. This is total impedance.
to get value of XL, I used (Z)square= (R)square+ (XL)square
XL, I got is 96.93
139.27^2=100^2+XL^2
XL^2=139.27^2-100^2
XL=96.93 ohms
I agree
using Xl=2*pie*f*L
I got L= 0.000771 =771uH
96.93=wL ; w=f*2*pi = 20e3*2*pi = 125.664e3 rad/s
96.93=125.664e3*L
L=96.93/125.664e3 = 0.000771 H
L=771uH
Excellent. You got the correct answer.
Now go back and plug it that value, solve for circuit impedance and verify that the impedance and current end up correct using that value. With homework you have time to do this. If on a test I would first finish the test and then if you have extra time left during the test you can go back and plug in your solution and verify it all works out.
is this correct?
Yep.
I disagree:Your final answer is out by a factor of 1000
How many uH are there in 1 Henry ?
Les.
by Ikimi .O