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

 

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.

 

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.

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

 

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.

 

M

At t=0 the value of the exponential function is 1, and 1 -1 = 0 so that is t.

How to derive formula for current that is flowing in circuit

 

This is trivial if you know calculus and differential equations. Do you?

 

This is trivial if you know calculus and differential equations. Do you?

My attempt

 

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.

 

You're close. Can you solve: e - IR - L•dI/dt = 0 for I = ƒ(t)?
.

What do you mean by e, I think voltage correct. Please confirm

My mistake small i and capital I are same things

 

What do you mean by e, I think voltage correct. Please confirm

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

 

- dt/(L/R)
Integrate

By taking integration answer is - (R /L) *t

 

Recheck your last step.
Then think about what Imax is.

 

Recheck your last step.
Then think about what Imax is.

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

 

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)]

 

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)]

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]?

 

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.

 

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.

I looked that is good. Also I have equation for i(t) given in text book. and I can find out voltage across resistor and inductor. Look at attached page