# inductors

Discussion in 'Homework Help' started by lemon, Feb 8, 2010.

1. ### lemon Thread Starter Member

Jan 28, 2010
125
2
A circuit comprises an inductor of 700mH of negligible resistance connected in series with a 15Ω resistor and a 24V d.c. source. Calculate:

a) i) the initial rate of change of current ii)the time constant (T) for the circuit
b) the value of the current after 1T seconds, and ii) the time to develop maximum current.

The attempt at a solution
The inductor has negligible resistance so we can use V=IR to calculate the current through the circuit = 24/15 = 1.6A
V-iR/L = 24-(1.6x15)/(700x10-3)=0

So, I got zero. So, I got nothing, right?
What am I doing wrong?
I suck at this!

2. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Write down your try step by step and indicate for which question is each part.

3. ### lemon Thread Starter Member

Jan 28, 2010
125
2
I could't work out in the editor how to show something raised to a power. Hope you can understand this easily.

a)i) So, the initial rate of change of current is when di/dt=V/L
Which is 24V/(700x10^-3)=34.3A/s (3s.f.)

ii) the time constant (T) for the circuit
This is not given in the same way as the time constant for a series connected CR.
Therefore T=(700x10^-3)/15Ω=0.0467s or 0.05s (2s.f.)

b) the value of the current after 1T seconds
Well there are two equations here - one for growth of current and one for decay.
I think it is growth because the question says 'initial rate'
Growth=I(1-e(-t/T))=34.3(1-e^(-0.05/0.05))=21.68A (2s.f.)
But - Currents will be approximately 63% or their final values after 1T when growing.
Imax=V/R=24/15=1.6A
So, at 1T, I=0.63x1.6=1.008A
These figures don't agree

ii) the time to develop maximum current. When t≥5T.
Imax=V/R=24/15=1.6A

4. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Another way is the same way calculators do it, 2.2E4 is $2.2X10^4$.

There is another way, I'm trying to figure it out, using LaTeX.

$2.2X10^4$

Use {tex][/tex}with the number between them, and brackets [] around tex and /tex.

A how to guide for math on this forum...

Last edited: Feb 13, 2010
5. ### mik3 Senior Member

Feb 4, 2008
4,846
63
You have to put I=Imax in the equation and not the rate of change.

6. ### mik3 Senior Member

Feb 4, 2008
4,846
63
To find the time rearrange i=Imax(1-e(-t/T))

as to get t=-T*ln(-i/Imax+1)

when i=Imax, ln(-1+1)=ln(0)=infinity

Thus, the time to reach Imax is infinite. This the reason it is said that Imax is reached after 5T.

7. ### lemon Thread Starter Member

Jan 28, 2010
125
2
t=-T*ln(-i/Imax+1)

Well. I had two values of i (not sure which one was correct)

t=-0.05*ln(-1.008/1.6+1)=i get mathematical error

or

t=-0.05*ln(-21.68/1.6+1)=i get mathematical error

8. ### mik3 Senior Member

Feb 4, 2008
4,846
63
In the first case you shouldn't get a mathematical error however the value for i is wrong. It should be i=Imax.

In the second case you will get a mathematical error because you have a negative number in the ln() which is not allowed.

9. ### lemon Thread Starter Member

Jan 28, 2010
125
2
if i=Imax then i=1.6A

t=-0.05*ln(-1.6/1.6+1)

but i'm confused cause they both have a minus sign in the ln(-1.008) and ln(-21.68)

but which one of these values is also correct from the earlier workings?

b) the value of the current after 1T seconds
Well there are two equations here - one for growth of current and one for decay.
I think it is growth because the question says 'initial rate'
Growth=I(1-e(-t/T))=34.3(1-e^(-0.05/0.05))=21.68A (2s.f.)
But - Currents will be approximately 63% or their final values after 1T when growing.
Imax=V/R=24/15=1.6A
So, at 1T, I=0.63x1.6=1.008A
These figures don't agree

10. ### mik3 Senior Member

Feb 4, 2008
4,846
63
ln(-1.008/1.6+1)=ln(0.37)

ln(-21.68/1.6+1)=ln(-12.55) this is wrong because 21.68 is wrong

The correct value of current after T is 1.008.

11. ### lemon Thread Starter Member

Jan 28, 2010
125
2
t=-0.05*ln(1.008/1.6+1)=0.05 (2d.p.)

I should not put the minus sign for i=Imax?

12. ### mik3 Senior Member

Feb 4, 2008
4,846
63
You have to put the minus for i unless you will get a negative time value which is not valid.

13. ### lemon Thread Starter Member

Jan 28, 2010
125
2
ln(-1.008/1.6+1)=ln(0.37)

but -1.008/2.6=-0.387...

and you don't have a minus, ln(0.37)

14. ### mik3 Senior Member

Feb 4, 2008
4,846
63
First, you have to make the division and then the addition, simple maths.

It is ln[(-i/Imax)+1]

15. ### lemon Thread Starter Member

Jan 28, 2010
125
2
right. -1.008/1.6 + 1=0.37

ln(0.37)=-0.9942....
-1s (2s.f.)

16. ### lemon Thread Starter Member

Jan 28, 2010
125
2
hold on!

To find the time rearrange i=Imax(1-e(-t/T))

as to get t=-T*ln(-i/Imax+1)

when i=Imax, ln(-1+1)=ln(0)=infinity

Thus, the time to reach Imax is infinite. This the reason it is said that Imax is reached after 5T.

So, the answer to this part of the question is infinity?
It can't be -1s. It doesn't make any sense for the current to be maximum at -1s, the circuit isn't connected before time t=0.

17. ### mik3 Senior Member

Feb 4, 2008
4,846
63
If you calculate the time mathematically, then (t) is infinity. However, if you assume that (i) reaches Imax after 5T, then t=5*0.05=0.25.

18. ### lemon Thread Starter Member

Jan 28, 2010
125
2
ahh! I see. So the answer to my question is:
(i) reaches Imax after 5T, then t=5*0.05=0.25.

But, you have been showing me how to calculate the time mathematically, then (t) is infinity?

But why is my answer -1s?

19. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Yes, the exact time is infinity however we assume that after 5T it reaches Imax because (i) is very close to Imax.

How you get -1?

20. ### lemon Thread Starter Member

Jan 28, 2010
125
2
How you get -1?

It's a good question. I must have eaten a strange fruit this morning. I will use 0.25s then. Once again mik3, thanks for all your great support today. It is very appreciated.