Problems figuring out this inductance formula

[johnboyman]

TL;DR Summary

I have questions about the I= 1/L∫Vdt formula

Hello. I am working this formula. I= 1/L∫Vdt . The following website have the best explanation of it.

http://www.learningaboutelectronics.com/Articles/Inductor-current-calculator.php

I am confused with a few things about it. One is that this website says If the current is Ac then the value will be a sine or a cosine waveform. How do I know which one to choose. I don't want to just pick one at random.

This example on this site looks like the following. This equation switched from cos to sin and when it does that's where it looses me.
What is the current flowing across an inductor if the voltage is 5cos(60t) and the inductance is 5H?

V= 1/L∫Vdt= (1/5H)∫(5cos(60t))= (5/300)sin(60t) A

So the current flowing across the inductor is (5/300)cos(60t) A.

I do not understand the steps that were taken to get this this final result. I hope someone has some advice thanks.

 

[DaveE]

Have you studied any calculus yet?

The example you cited looks good to me. Honestly, I don't really know how to simplify the solution beyond:

1) I = (1/L)⋅∫v(t)⋅dt
2) v(t) = 5⋅cos(60t), L = 5
=> I = (1/5)⋅∫5⋅cos(60t)⋅dt = ∫cos(60t)⋅dt = sin(60t)/60

 

[johnboyman]

I have studied some calculus.
Where did (1/5)⋅∫5⋅ go in the beginning of the equation and in the end how did you go from cos to sin and where did /60 come from? Sorry to to ask, i have gaps in my math.
I = (1/5)⋅∫5⋅cos(60t)⋅dt = ∫cos(60t)⋅dt = sin(60t)/60

 

[DaveE]

 

[DaveE]

For now, until you have studied more math, I would just memorize the formulas:
∫sin(ω⋅t)⋅dt = -(1/ω)⋅cos(ω⋅t) and ∫cos(ω⋅t)⋅dt = (1/ω)⋅sin(ω⋅t).

To learn more about these formulas, I would look for explanations of integral calculus. Khan Academy or 3Blue1Brown are good sources.

 

[johnboyman]

Thanks alot

 

[DaveE]

There is an alternate formula for how an inductor works, which I personally prefer. It is v(t) = L⋅(di(t)/dt) where di/dt is the derivative of the current. The derivative is the rate of change, or the slope (vs. time) of the current.

It's just two different ways of explaining the same physical phenomenon, so don't worry if it's confusing. You may prefer one version vs. the other.

 

[Tom.G]

=> I = (1/5)⋅∫5⋅cos(60t)⋅dt = ∫cos(60t)⋅dt = sin(60t)/60

Yeah, confusing. The change from "∫cos(60t)⋅dt" to "sin(60t)/60" works because sin and cos are integrals, and derivitives, of each other.

You can see that if you sketch a sin wave and then sketch a cos wave below it. You will note that the inflection point (peak) of one is at the Zero crossing of the other.

Cheers,
Tom