# Hie, Can anyone tell me the formula for finding Inductor value in this circuit? In advance Thanks

#### DeepThinker

Joined Nov 26, 2018
24

#### Alec_t

Joined Sep 17, 2013
12,248
Is the 718mV peak-to-peak or rms?
What load will it be driving?
Is this a school/college problem?

• DeepThinker

#### DeepThinker

Joined Nov 26, 2018
24

#### DeepThinker

Joined Nov 26, 2018
24
Yes, 718mV is (p-p). Can you please tell me the formula or steps to find the inductor value in this circuit

#### ericgibbs

Joined Jan 29, 2010
13,825
hi DT,
Consider what would the resistance offered by the Inductor have to be to reduce 0.5Vpk t0 0.359Vpk.

E

#### DeepThinker

Joined Nov 26, 2018
24
hi DT,
Consider what would the resistance offered by the Inductor have to be to reduce 0.5Vpk t0 0.359Vpk.

E
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 inductance

#### LesJones

Joined Jan 8, 2017
3,506
What is the formula for the REACTANCE of the inductor ?
Are you allowed to assume the resistance of the wire in the inductor is zero ?
Do you know how to do vector addition of resistance and reactance ?
I do not agree with your calculation you for the resistance of the inductor treating it as it was pure resistance.

Les.

• click_here and ericgibbs

#### AlbertHall

Joined Jun 4, 2014
11,524
Look here and scroll down to 'AC Supply through an LR Series Circuit'

#### dcbingaman

Joined Jun 30, 2021
442
You know the current in the circuit. It is given by the output and the resistor. I am assuming this is an 'ideal' inductor'. That is an inductor with zero series resistance. Otherwise you do not have enough information to solve the problem. You would need to know the phase angle of the output voltage. This is just a voltage divider just remember you need to work with complex numbers to get the answer. So you know the voltage across the inductor and you know the current along with the frequency. Keep in mind the inductor will be purely reactive at jwL reactance. Take it from there.

• DeepThinker

#### dcbingaman

Joined Jun 30, 2021
442
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).

• DeepThinker

#### MrAl

Joined Jun 17, 2014
8,496
Since 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.

• DeepThinker

#### DeepThinker

Joined Nov 26, 2018
24
Since 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.
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

#### Papabravo

Joined Feb 24, 2006
16,938
If you replace the inductor with a resistor and solve the problem for any and all frequencies, what value would that be?
What inductance would you need at 20 kHz to replicate an impedance equivalent to that resistance?

• dcbingaman

#### LesJones

Joined Jan 8, 2017
3,506
You 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.

• dcbingaman and DeepThinker

#### DeepThinker

Joined Nov 26, 2018
24
You 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.
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?

Last edited by a moderator:

#### LesJones

Joined Jan 8, 2017
3,506
How many uH are there in 1 Henry ?

Les.

#### ericgibbs

Joined Jan 29, 2010
13,825
hi DT.

Ask if you have a query
E

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• DeepThinker

#### dcbingaman

Joined Jun 30, 2021
442
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.

#### LesJones

Joined Jan 8, 2017
3,506
DeepThinker, You were correct. It was me that was wrong.

Les.

• DeepThinker

#### dcbingaman

Joined Jun 30, 2021
442
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