initial rate of growth of the current:

Thread Starter

ironasona

Joined Nov 12, 2020
53
I just had a question how would i calculate the initial rate of growth of the current

Specifications
Voltage source - 100v dc
Coil having a resistance of 10ohms and inductance of 5H

Which formula is right
is it

100v/10ohms = 10A/s or is it 100v/5H =20A/s its 100v/5ohms = 20A/s right ?
 

MrChips

Joined Oct 2, 2009
24,430
None of them are correct.
You must have dimensional equality in your calculations, i.e. units on the left hand side must be equal to the units on the right hand side.

Volts/Ohms = Amps
You cannot equate Volts/Ohms with Amps/seconds.

Similarly Volts/Henries is never equal to Amps/seconds.
Edit: oops. Pb is right. Volts/Henries = Amps/seconds
 

Papabravo

Joined Feb 24, 2006
17,001
The operative formula comes from the differential equation for the series LR circuit. If the R was zero you would be (almost) correct, but since it is there you have to work a bit harder

\( V_S\;=\;L\cfrac{di}{dt}\;+Ri \)

The solution to which is:

\( i(t)\;=\;\cfrac{V_S}{R}\left(1\;\boldsymbol{-}\;e^{\boldsymbol{-}\left(\tfrac{R}{L}t\right)}\right) \)

The initial value of the current at t=0 is 0 because the exponential term evaluates to 1, and 1 - 1 is still zero, even after all these years. What we wanted was the initial value of the derivative. We have a constant out front which stays but the derivative of what is in the parentheses is:

\( \cfrac{di}{dt}\;=\;\cfrac{V_s}{L}e^{\boldsymbol{-}\left(\tfrac{R}{L}t\right)} \)

The growth is exponential. and the initial rate at t = 0, as you said Vs/L = 20 A/sec

Edit: I had a hard time getting the LaTex to be readable; \boldsymbol{-} was necessary to get the - sign in the exponent to be visible. Sheesh!

Edit 2: The subscript on the voltage V should be "S" for the 100 V source driving the circuit
 
Last edited:

RBR1317

Joined Nov 13, 2010
668
Don't need those fancy schmancy equations anyway. (And you still need to take the first derivative of I(t) to get the growth rate.)

Just realize that the initial growth rate of a simple exponential curve will reach the final value in one time constant.
 

MrAl

Joined Jun 17, 2014
8,502
The operative formula comes from the differential equation for the series LR circuit. If the R was zero you would be (almost) correct, but since it is there you have to work a bit harder

\( V_L\;=\;L\cfrac{di}{dt}\;+Ri \)

The solution to which is:

\( I(t)\;=\;\cfrac{V}{R}\left(1\;\boldsymbol{-}\;e^{\boldsymbol{-}\left(\tfrac{R}{L}t\right)}\right) \)

The growth is exponential. and the initial rate at t = 0, is V/R = 10 A/sec.

Edit: I had a hard time getting the LaTex to be readable; \boldsymbol{-} was necessary to get the - sign in the exponent to be visible. Sheesh!

Hi there,

I think you may have made a mistake because of having problems with the math software. Math software has always been a problem on the web for some reason or another things often go wrong that's why i mostly use pure text like y=2*x+1.

For one, your DE does not include the source voltage of 100vdc only the inductor voltage:
VL=L*di/dt+R*i

but if we change that VL to the source voltage V which is 100vdc when turned on, we get a simple expression:

V=L*di/dt+R*i
divide by L to isolate di/dt and get:
V/L=di/dt+R*i/L
get di/dt to one side:
di/dt=V/L-R*i/L
then substitute actual values:
di/dt=100/5-10*i/5
(and note that i is i(t) and that is the inductor current)

and we end up with a DE that is clearly the slope of the current. Now given that the initial current is zero (the initial current not the initial slope) we set i=0 and get:
di/dt=100/5

which of course equals 20 amps per second.

Now if that is true, then if we multiply 20 by some small time value like 0.001 seconds, then we should see a graph of the entire solution for the current reach the point 20*0.001=0.020 amps in 1ms. That could be a check point.
 

Papabravo

Joined Feb 24, 2006
17,001
Hi there,

I think you may have made a mistake because of having problems with the math software. Math software has always been a problem on the web for some reason or another things often go wrong that's why i mostly use pure text like y=2*x+1.

For one, your DE does not include the source voltage of 100vdc only the inductor voltage:
VL=L*di/dt+R*i

but if we change that VL to the source voltage V which is 100vdc when turned on, we get a simple expression:

V=L*di/dt+R*i
divide by L to isolate di/dt and get:
V/L=di/dt+R*i/L
get di/dt to one side:
di/dt=V/L-R*i/L
then substitute actual values:
di/dt=100/5-10*i/5
(and note that i is i(t) and that is the inductor current)

and we end up with a DE that is clearly the slope of the current. Now given that the initial current is zero (the initial current not the initial slope) we set i=0 and get:
di/dt=100/5

which of course equals 20 amps per second.

Now if that is true, then if we multiply 20 by some small time value like 0.001 seconds, then we should see a graph of the entire solution for the current reach the point 20*0.001=0.020 amps in 1ms. That could be a check point.
I did use the wrong subscript. It should have been the driving voltage of the source.
 

MrAl

Joined Jun 17, 2014
8,502
I did use the wrong subscript. It should have been the driving voltage of the source.
Oh yes and i bet it had something to do with playing around trying to get a decent Latex graphic from the equation you wanted to show.

It's too bad this kind of thing happens. It happens with other things on the web too that were not well thought out beforehand i guess, or else just a little more complicated to see ahead with.
Another striking example is the lower case L and the number 1, and especially the lower case L and the upper case "i" in some font types.

The only solid method i have found that works across nearly all platforms is a graphic upload. Using a gif or png or jpg shows equations really well. Just hopefully they dont get lost somehow on the server :)
 

Papabravo

Joined Feb 24, 2006
17,001
Vs/R doesn't have the right units. It should be Vs/L = 20 A/sec.
I thought that both my sources had had the same error. I might as well give up and retire. The initial value of current is given by the first expression and you have to take the derivative to get the initial value of the growth rate.
 
Last edited:

MrAl

Joined Jun 17, 2014
8,502
I thought that both my sources had had the same error. I might as well give up and retire. The initial value of current is given by the first expression and you have to take the derivative to get the initial value of the growth rate.
Now you know why i dont bother with Latex too much. You end up putting more effort into getting it to show up right and pay more attention to that then the actual content and it's distracting from the main goal.
Now if someone had a really really really really good Latex editor maybe it would work, maybe :)
I would say we would need a text to Latex converter with no BS'ing around. You enter:
y=x^2+x+1
and it spits out the exactly correct Latex representation and ALL web sites use the same Latex format.
 

Papabravo

Joined Feb 24, 2006
17,001
I find it easier to use MS Office and take a screen grab.
If it is more than a line or two I would agree. Libre Office has a very fine math package. IIRC Mr. Softie used to charge for the add on (ca. 2002) which is why I gave them the heave-ho.
 

Papabravo

Joined Feb 24, 2006
17,001
Note that the vertical scale on the right hand side of the plot has the correct value and units for the derivative, which is actually a finite difference derivative rather than a closed form computation. It is great for measuring gain, slew rate and other slopes.
 

MrAl

Joined Jun 17, 2014
8,502
For your interest, below is the LTspice simulation of the circuit:
Note the L/R time-constant value.

View attachment 244768
I meant to graph that too and then show the sanity check i talked about in my post.
I used your graph instead.

In the following graph the point (0.1,2) is indicated but for a more accurate approximation we would use the point (0.001,0.020) or even a shorter time value like 1us.
Using the point (0.1, 2.0) we get just an approximation as this is a numerical approximation and that is because on a graph like this the x value would be slightly different than 0.1 to get exactly 2.0 on the plot, or stated differently with a value of x=0.1 the graph would not be at exactly 2.0 it would be slightly different. Using x=0.001 or x=0.000001 results in a more accurate approximation but this is good enough.

RL_CircuitDVDT-2.png
 
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