Problems figuring out this inductance formula

In summary, the conversation discussed a formula for calculating the current across an inductor and a website with a good explanation of it. There was confusion about the use of sine or cosine waveform and the steps taken to simplify the equation. The conversation also mentioned an alternate formula for understanding how an inductor works. The explanation of the integration and derivation of sine and cosine was suggested for further understanding.
  • #1
johnboyman
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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.
 
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  • #2
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
 
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  • #3
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
 
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  • #5
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.

 
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  • #6
Thanks alot
 
  • #7
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.
 
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  • #8
DaveE said:
=> 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
 
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Related to Problems figuring out this inductance formula

What is the inductance formula and how is it used?

The inductance formula is used to calculate the amount of inductance in an electrical circuit. It is represented by the symbol L and is measured in henries (H). The formula is L = N^2 * μ * A / l, where N is the number of turns in the coil, μ is the permeability of the material, A is the cross-sectional area of the coil, and l is the length of the coil.

What are some common problems encountered when using the inductance formula?

Some common problems encountered when using the inductance formula include incorrect values for any of the variables, using the wrong units for measurement, and not taking into account the effects of external magnetic fields. It is important to double check all values and units and to consider any external factors that may affect the inductance calculation.

How do I solve for inductance when given the other variables?

To solve for inductance when given the other variables, simply plug in the known values into the inductance formula and solve for L. It is important to ensure that all values are in the correct units and to use the proper mathematical operations.

Can the inductance formula be used for all types of circuits?

The inductance formula can be used for most types of circuits, including simple circuits with just a coil and more complex circuits with multiple components. However, it may not be accurate for circuits with non-linear elements or for circuits with high frequencies.

Are there any alternative formulas for calculating inductance?

Yes, there are alternative formulas for calculating inductance, such as the mutual inductance formula for circuits with multiple coils and the self-inductance formula for calculating the inductance of a single coil. These formulas may be more appropriate for certain types of circuits or for specific calculations.

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