# Exponential function and equations

Discussion in 'General Electronics Chat' started by vead, Sep 5, 2016.

Nov 24, 2011
621
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Hello
Exponential function that take argument any return value of e raised to power x
F(t) =e^t
F(0)=1,f(1)=2.718, f(2)=7.38,f(3)=20....etc
It can be seen function is raising exponentials. I understand exponentials function. But I am having problem how they relates to electronics. How voltage and current signal represent as exponential function. So I trt to understand, look below example charging current flowing in series RL circuit as exponentials function over time

2. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
I'm not sure what your question is. An exponential function with a negative exponent will asymptotically approach 0 as time goes to ∞. If you subtract an exponential function with a negative exponent from a constant, like 1, then the result will approach 1 as time goes to ∞. An exponential function with a positive exponent represents a system that is unstable and has the potential to destroy itself.

Nov 24, 2011
621
8
Have you seen the formula in last line? There is formula for current, that flow in RL series circuit. It is exponentials function. I don't understand that formula. I can solve questions but I don't understand concept behind the formula

4. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
At t=0 the value of the exponential function is 1, and 1 -1 = 0 so that is the value of the current at t=0. At some large value of t the value of the exponential is very nearly equal to 0, and so the value of 1 - ε, where ε is a very small number, is just a little less than 1 which is the current at that very large value of t.

Nov 24, 2011
621
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M
How to derive formula for current that is flowing in circuit

6. ### wayneh Expert

Sep 9, 2010
12,392
3,246
This is trivial if you know calculus and differential equations. Do you?

Nov 24, 2011
621
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My attempt

8. ### wayneh Expert

Sep 9, 2010
12,392
3,246
You're close. Can you solve: e - IR - L•dI/dt = 0 for I = ƒ(t)?
Consider the boundary conditions when the switch opens or closes, when t=0 or t=infinity

Hint: Separate the dI and I, and dt terms so you can integrate them separately.

Nov 24, 2011
621
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What do you mean by e, I think voltage correct. Please confirm

My mistake small i and capital I are same things

Last edited: Sep 5, 2016
10. ### wayneh Expert

Sep 9, 2010
12,392
3,246
Yes, e or emf. Note my previous hint.

dI/dt = (V - IR)/L = (V/R - I)/(L/R)
dI /(I - V/R) = - dt/(L/R)
Integrate

Nov 24, 2011
621
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By taking integration answer is - (R /L) *t

Last edited: Sep 5, 2016
12. ### wayneh Expert

Sep 9, 2010
12,392
3,246
Then think about what Imax is.

Nov 24, 2011
621
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My mistake I=V/R+e^_(R/L) *t ampear
At t=0
I(0)=V /R amp
At t=infinity
I=V/R

Last edited: Sep 5, 2016
14. ### wayneh Expert

Sep 9, 2010
12,392
3,246
Right, so you can rearrange the result to:
I(t) = I(0) • [1 -exp(-t•R/L)]

You can further define the time constant tau as L/R, giving:
I(t) = I(0) • [1 -exp(-t/tau)]

Nov 24, 2011
621
8
I am confused,
We have equations
I(t) =V/R +e ^-(R /L) *t
How it become I(t) =I(0).[1-e ^-(R /L)*t]?

16. ### wayneh Expert

Sep 9, 2010
12,392
3,246
My bad. You had an earlier error I missed. I was going to draw this out for you but I found a nice exposition of it online.

See here.