Time Constant Induction question

AI Thread Summary
The discussion revolves around a physics problem involving a coil's inductance and resistance in a D.C. circuit. The user seeks help calculating the resistance, energy stored, and time for current decay based on given parameters. Key equations for current rise and time constant are discussed, with emphasis on the relationship between current, inductance, and resistance. Participants clarify the correct use of natural logarithms in calculations and confirm that the time constant is approximately 2.88 ms. The conversation highlights the importance of accurately transposing equations to solve for unknown variables.
Wayne
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Hi all, I need your help with a Time constant question? I am studying for an exam via distance learning and don't feel they have covered this question, question taken from past paper.

2. A coil of inductance 2 H and unknown resistance is connected to a D.C. supply of
100 volts. After 4 ms the current has risen to 75% of its final steady state value.
Calculate EACH of the following:
(a) the resistance by the coil;
(b) the energy stored in the coil when the current has reached its steady state value;
(c) the time taken for the current to fall to 50% of its steady state value when the supply
is switched off.

The question I am stuck on is a, If I can find out what 75% is I can then find out full current and time constant to to transpose, tc=L/R?

I understand 1 tc is 63.2% and 5 tc is 100%

Thank you in advanced.

Wayne
 
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Hi Wayne. Welcome to Physics Forums. :wink:

Can you write the full equation that describes the rise of the current vs. time?
 
i = I (1-e (-t/tc))
i =instantaneous current
I final current I=v/r
 
So at t=4ms equate this to 75% of If.

You know that value of the final current, If.
 
Sorry you lost me! I don't know i as not given and can't calculate I as don't know r to calculate I=v/r
 
but they do tell you that 'i' is 75% of 'I'. So looking at your equation i = I (1-e (-t/tc)) Can you rearrange, so that you don't actually need to find 'i' and 'I' separately?
 
Sorry I totally not getting this been looking at this for a week before I decided to put it on here.

I understand i is 75% of I and took 4 ms, and guessing I have to us e as the curve is expo-dental 1 tc=63.5% &
5 tc is 100%
 
Ok has the penny just dropped?

if formula i = I (1-e (-i/I))

Is transposed to

if tc = t / ( -in ((1- (i/I)))

i= 0.75 & I =1

Then tc = t / (-in ((1- (0.75/1))) tc= 1.68 ms
 
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I think you are almost on the right track now. But your equations are a little hard to read. I think you might have made a couple of typing mistakes... I'm guessing for the first formula you meant i = I (1-exp(-t/tc)) and I'm guessing that 'in' means natural logarithm. Then I would say your final equation tc = t / (-in ((1- (0.75/1))) is correct. But I don't get 1.68ms when I solve for tc ...
 
  • #10
Opps slight miss calculation
tc=2.88ms?
"-in" means natural logarithm
 
  • #11
yeah :) I got tc=2.88ms too. nice. p.s. I think there should also be a negative sign in the calculation somewhere. But 2.88ms is the right answer, so I guess you did remember to use the negative sign in there when you did the calculation.

edit: or actually, I got 2.89ms, you might want to check the rounding of the number
 
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  • #12
Wayne said:
Ok has the penny just dropped?

if formula i = I (1-e (-i/I))

Is transposed to

if tc = t / ( -in ((1- (i/I)))

i= 0.75 & I =1
This should be i=0.75I, so the exponent becomes 0.75I/I. Then the I/I cancels.

I'm not sure why you are using "in" for "ln" but the latter is definitely recommended. :wink:

Looks like BruceW has put you right now.
 

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