LR Natural Response circuit

[cartman007]

Hi Everyone.

I have what seems to be an easy LR circuit problem I would like to have some input.



Is it safe to assume that because first 2K-Ohm resistor is actually in parallel with the 8K-Ohm resistor since the inductor at 0 is a short?
Also will the 2K+4.7K = 6.7K also be in parallel with the inductor and the rest of the resistors?

Example:


OR will the 6.7K-ohm just be omitted?

 

[c_mitra]

It is correct to assume that 4.7K and 2K are in series and can be replaced with a 6.7K. This 6.7K can be put as parallel to the inductor.

It is wrong to assume that the inductor is a short at t=0.

The next two diagrams are not the same as the first one.

You need to set up the proper equations and solve them.

 

[cartman007]

Rtot = ( 1/8k + 1/2K)^-1 = 1.6k ;
L = 10x10^-3H ;
Tao = 10x10^-3/1600 = 6.25x10^-6 seconds ;
V|| = 24V ;
i(t) = Ioe^(-t/Tao) ;
Io = 24/1600 = 0.015 ;
i(t) = 0.015e^(-t/6.25x10^-6)
i(t) = 0.015e^(-160000t)

Soo at i(10) = 0.015e^(-160000(10)) = 0 .... is this right?

 

[FvM]

No. Equivalent source resistance for L/R time constant calculation is 2 k || 6.7 k = 1.54 k. Equivalent source voltage is 24V * 6.7/(2 + 6.7) = 18.48 V

 

[c_mitra]

...Soo at i(10) = 0.015e^(-160000(10)) = 0 .... is this right?

NO!!

It is a simple circuit: you need to set up the equation as per diagram. Indicate currents in the diagram (i1, i2 etc etc)

Indicate voltages as per their resistance or reactance. It will come as a differential equation (because of L; it would be same for C also).

It is good to learn the basics the traditional way.

 

[CataM]

OR will the 6.7K-ohm just be omitted?

The only thing you can omit in that circuit is the 8k resistor in order to calculate the current through the inductor. Nothing else.
Afterwards, to calculate the total current from the source, just add 24/8k amps.

 

[RobertFalbo]

I would start by, removing the 8 K, like they say above, add 3 ma (24 volts/8K ohms) back in later.

add the 2K and the 4.7K to give you a 6.7K

Now look at the voltage divider you have with the 2 K and the 6.7K resistor on 24V

V= 24V * 6.7K/(2K+6.7K) = 18.483..V

calculate the impedance of your 18.483 v

2K||6.7K, (2K*6.7K)/(2K+6.7K) = 1.54k... ohms

So now you have a 10 mH inductor in series with 1.54K ohm resistor connected to a 18.483 volt source.


so using the standard formula

**broken link removed**
https://www.electronics-tutorials.ws/inductor/lr-circuits.html



V/R= 18.483/1.54K =12 ma

-R/L = - 1.54K ohms / 10mH= -154K
@10 us
we get 12ma *(1-e^-10us*154K)=12 ma * .78567= 9.428 ma answer to 4.3

So now we take the 9.428 ma and calculate the IR drop of the 1.54k ohm impedance . 1.54K * 9.428 =14.52 volts.

So VL@10us = 18.483V- 14.52 v = 3.961 Volts answer to 4.1



I let somwbody else do the last two.
I got to go have breakfast

 

[c_mitra]

I let somwbody else do the last two.
I got to go have breakfast

Your results are approximate. You need to redo.

 

[RobertFalbo]

Your results are approximate. You need to redo.

yea, I only copied a few digits from the calculator. but used the full precision of the calculator during operations.

 

[c_mitra]

One poster with sharp eyes has correctly pointed out that the current through the 8K will be const and can be left out for the time being (you can add the const to the total current later).

Let the current through the inductor is i1(t) and the current through (2k+4.7k=6.7k) is i2(t). The current through the 2k is i1(t)+i2(t).

The voltage across the 6.7k is i2(t)*6.7k and it is the same across the inductor. Hence L*(di1/dt)=i2(t)*6.7k

We need another equation because we have two unknowns: 24=(i1(t)+i2(t))*2k+i2(t)*6.7k; put i2(t) in terms of i1(t); we get i2(t)=(24-i1(t)*2k)/8.7k

Put this in the previous equation: L(di1/dt)=6.7k*(24-i1(t)*2k)/8.7k; L=1/100; di1(t)/dt=100*6.7k(24-i1(t)*2k)/8.7k

Now you can calculate i1. this is of the form y'=a-b*y and can be solved easily. You need to put the boundary conditions.

 

[RobertFalbo]

One poster with sharp eyes has correctly pointed out that the current through the 8K will be const and can be left out for the time being (you can add the const to the total current later).

Let the current through the inductor is i1(t) and the current through (2k+4.7k=6.7k) is i2(t). The current through the 2k is i1(t)+i2(t).

The voltage across the 6.7k is i2(t)*6.7k and it is the same across the inductor. Hence L*(di1/dt)=i2(t)*6.7k

We need another equation because we have two unknowns: 24=(i1(t)+i2(t))*2k+i2(t)*6.7k ; put i2(t) in terms of i1(t); we get i2(t)=(24-i1(t)*2k)/8.7k

Put this in the previous equation: L(di1/dt)=6.7k*(24-i1(t)*2k)/8.7k; L=1/100; di1(t)/dt=100*6.7k(24-i1(t)*2k)/8.7k

Now you can calculate i1. this is of the form y'=a-b*y and can be solved easily. You need to put the boundary conditions.

I like your approach, but something seems off.

so now we take an integral from t=0 to t=10us (di/dt)

6.7k/8.7k *(24-i1(t) *2k) dt

100*(6.7/8.7) * 24t-1K*i1(t)^2 from t=0 to t=10us

I don't see the exponential expression?

 

[c_mitra]

I like your approach, but something seems off.

... ...

I don't see the exponential expression?

Your integration is wrong!

di1(t)/dt=-(b*i1(t)-a) where b=100*6.7k*2k/8.7k and a=24*100*6.7k/8.7k

d(i1)/(b*i1-a)=-dt

Now integrate:

ln(b*i1-a)=-bt+C ; C is the integration const

Now you can see the exponential form? Now I am feeling sleepy!!

 

[RobertFalbo]

Your integration is wrong!

di1(t)/dt=-(b*i1(t)-a) where b=100*6.7k*2k/8.7k and a=24*100*6.7k/8.7k

d(i1)/(b*i1-a)=-dt

Now integrate:

ln(b*i1-a)=-bt+C ; C is the integration const

Now you can see the exponential form? Now I am feeling sleepy!!

Thanks, I see where I went wrong, i1(t) is a function and not a variable.