Need Help finding inductance in a RL circuit

[chris2016]

Hi Everyone

I need some help finding the inductance of a RL circuit

All the information I have is that the resistor in the circuit is 1.500 k ohm

input to the RL circuit is 1 volt 80 khz

output 0.7 volts 80 khz

I have no information for the inductor

Every equation I find needs either henry or XL to calculate

The question asked is determine the inductance value of the RL circuit

Am I missing something to proceed ?

please Help

 

[Alec_t]

Welcome to EP!

Am I missing something to proceed ?

We're missing a schematic. Where is the output taken from?

 

[chris2016]

Hi Alec

Thanks for the quick reply

I can do the other equations ok but this really has me stuck

 

[Alec_t]

There are two possible configurations of R and L within your 'RL Filter' module, so there are two possible answers to your question.
This simulation demonstrates that a single inductance value would give two different results depending on the configuration.

 

[dorke]

Chris,
You haven't revealed all the information in the question.
Note that:
the resistance between 1-2 is 1.61K ohms and between 1-3 is 1.5K

Assuming this is a 1st order R-L filter.
The circuit is this:
From the scope trace you found that
Vi/Vout=0.7 and f=80khz.

Use Vin/Vout=R2/[R2+(R1+XL1)] to calculate XL1
and then knowing f find L1.


EDIT:
Chris,A question for you :
Why did I only put a single possible option for the circuit and not two like @Alec_t did?
If you can answer that correctly ,give yourself a bonus for understanding...

 

[chris2016]

Hi Dorke & Alec


To try and calculate I have been using the same style layout that Alec provided but I'm really stuck as no figure is shown for the inductor (h) and the only way I can try and find the impedance is by using V = IZ , but when I use this equation things don't add up on the impedance triangle.

I think the reason you have only done one diagram is that the meter is measuring the circuits with with DC voltage which has no effect on the inductor making it more like a resistor

So when making a diagram the 2 components can be shown as resistors as the inductor is not repelling the AC current to make any difference to the resistance

This is my first time doing the theory side of things and I'm puzzled

 

[dorke]

1. I didn't wont to totally solve this for you so I wrote:
"Use Vin/Vout=R2/[R2+(R1+XL1)] to calculate XL1".

Since that doesn't help(You should have noticed you can't simply add r+X,that should be a vector addition)

Let's re-write it
Vin/Vout=R2/[(R1+R2)+jXL1]=R2/Z
were Z= [(R1+R2)^2 +(XL1)^2]^0.5 ,vector addition

thus
Z=R2*1/(Vin/Vout)
squaring the equation:
(R1+R2)^2 +(XL1)^2 =[R2*1/(Vin/Vout)]^2
thus:
(XL1)^2 =[R2*1/(Vin/Vout)]^2 - (R1+R2)^2
All the right-end side values are known to you.
In the left-end side XL1=2*pi*f*L1 ,f is known to you.
Solve to get L1.

2. No bonus,you didn't answer correctly(yet).
It is correct that measuring resistance with an ohmmeter is done in DC (i.e f=0Hz).
Thus we only measure the values of pure resistors while the reactance X ,of the inductor is zero and has no effect on the measurement.

But that is not why I only did a single diagram.
Note that L1 is in the "horizontal branch",I could put it on the "vertical branch" like Alec did.
Notice that it would still agree with the ohm measurements if I did that.
So,Why didn't I ?

 

[chris2016]

Hi Dorke

Here's another shot in the dark

Did you put the inductor on the horizontal branch because the first equation is to calculate XL1 by using the input and output voltage information from terminals 2 & 3

 

[dorke]

Nop,
Here is the hint for you,the only information available to solve this...

 

[chris2016]

Hi Dorke

I see now

The resistor is between terminals 1 and 3 1.500 k

The inductor is on the input pin 2 and the resistor is on GND pin 1 giving us a meter reading of 1.610 k between pins 1 and 2 (inductor + resistor )

When we look at the oscilloscope readings it makes it an RL lag circuit as the output voltage lags the input voltage which puts the inductor on the input terminal (2) as your drawing shows

There is no other way is can be as you have shown

 

[dorke]

Yes Chris you have got the bonus,
well done!

Note that if the inductor would have been on the "vertical arm" ,
it would have been a "Lead-circuit".

In terms of Filters we talk about phase-shift.

 

[chris2016]

Hi Dorke

I really appreciate all your help, can you check the equations below to see if I'm doing them correctly

Z=R2*1/(Vin/Vout)

R2 x 1 = 1.5 = 1.05 k
Vin 1v / Vout 0.7v = 1.428571429
-----------------------------------------------------------------------------------------------

(XL1)^2 =[R2*1/(Vin/Vout)]^2 - (R1+R2)^2

(XL1)^2 =[1.5/1.428571429]^2 - (1.61)^2
(XL1)^2 = 1.1025 - 2.5921
(XL1)^2 = -1.4896

Am I going in the right direction here?







-----------------------------------------------------------------------------------

 

[dorke]

The direction is fine, but some numeric corrections:

(XL1)^2 =[1500*1.428571429]^2 - (1610)^2
(XL1)^2 =(2143)^2- (1610)^2

 

[chris2016]

Hi Dorke

Can you check this for me

Z=R2*1/(Vin/Vout)

R2 x 1 =1500 = 1500
Vin 1v / Vout 0.7v = 1.428571429
Z= 1050
-----------------------------------------------------------------------------------------------

(XL1)^2 =[R2*1/(Vin/Vout)]^2 - (R1+R2)^2

(XL1)^2 =[1500*1.428571429]^2 - (1610)^2
(XL1)^2 =(2143)^2- (1610)^2
(XL1)^2 =4592449 - 2592100
(XL1)^2 = 2000349
XL1 = square root of 2000349
XL1 = 1414.33 ohms


phase shift = tan -1(XL/R)
= tan -1(1414.33/1610)
= tan -1(0.87846583)
= 41.29820109 radians


Still a little confused what to do to get L1 as below

In the left-end side XL1=2*pi*f*L1 ,f is known to you.
Solve to get L1.

 

[dorke]

from XL1=2*pi*f*L1
L1=XL1/[2*pi*f]=1414/[6.28*80000]=0.002815H

 

[chris2016]

Hi Dorke

Thats great

Thanks for all your help

 

[chris2016]

Hi Dorke

One last question I promise

For an RL lead circuit do the equations remain the same ? Z=R2*1/(Vin/Vout)

 

[dorke]

No, how could they be?
It is a different circuit,L1 in the vertical arm.

Vout/Vin=(R2+jXL1)/[(R1+R2)+jXL1]=(R2+JXL1)/Z

Note:
I see now that the notations only,in all the above are wrong!
It should be Vout/Vin like here.(=0.7)
But the calculations are correct.

So simply replace all those (vin/vout) with (Vout/Vin).
all the values and calculations remain the same.

 

[chris2016]

Hi Dorke

Thanks for pointing that out , I was wondering why the calculation wasn't matching when using this equation for Z = Square root R^2 + XL^2
I understand now that the line is crossing the horizontal axis of the phasor so the equation is slightly different and the vout is taken across the inductor


I know a need to re write the following equation but not sure how

Vout/Vin=(R2+jXL1)/[(R1+R2)+jXL1]=(R2+JXL1)/Z

RL Lead circuit details

Channel 2 input voltage 1 v

frequency 3.311 Hz

Channel 1 Output Voltage 0.7 v

 

[dorke]

Everything is the same:
Z is the same(assuming the same resistance measurements) , Vout/Vin=0.7,
f=3.311Hz

0.7= [(R2)^2 +(XL1)^2]^0.5 / [(R1+R2)^2 +(XL1)^2]^0.5
solve for XL1