Purely Inductive Circuit -- Mathematical proof for current lag

[CosmicC]

how we can mathematically prove that in a purely inductive circuit current lags behind voltage by a phase angle of π/2?

 

[BvU]

With a purely imaginary impedance $j\omega L$ and $V = Z \;I$ you see that $\arg V = \arg j\omega L + \arg I$

 

[berkeman]

how we can mathematically prove that in a purely inductive circuit current lags behind voltage by a phase angle of π/2?

Welcome to the PF.

1) Is this a question for your homework or schoolwork?

2) << EDIT -- I removed this point since it might not be accurate >>.

3) Are you familiar with the differential equation defining the voltage across an inductor v(t) as a function of the inductance and the derivative of the current i(t) through the inductor?

 

[cnh1995]

@CosmicC, see if this recent discussion helps.
https://www.physicsforums.com/threads/inductor-behaviour-in-an-ac-power-supply.921304/
The topic is a bit different from your OP, but it contains the math you are looking for.
Read #12 and #13.

 

[BvU]

Point isn't so much we don't want to help you; more that we have no idea what kind of help is useful for you. Are you at all familiar with complex numbers when dealing with AC, or with the equations berkeman is referring to ?
In short: a bit more context, please !

Oh, and: read the guidelines

 

[CosmicC]

Welcome to the PF.

1) Is this a question for your homework or schoolwork?

2) << EDIT -- I removed this point since it might not be accurate >>.

3) Are you familiar with the differential equation defining the voltage across an inductor v(t) as a function of the inductance and the derivative of the current i(t) through the inductor?

Yes Sir I am.

 

[BvU]

So we can mark this one as solved ? Or is there a remaining question ?

 

[jim hardy]

how we can mathematically prove that in a purely inductive circuit current lags behind voltage by a phase angle of π/2?

First you have to define voltage as some math function.

Your question infers sine function but doesn't say that's what it is. Sine is a mathematical oddity in that its derivative and integral have its same shape .
so we use them almost interchangeably

Once you realize that it's trivial

Yes Sir I am.

So you're aware e = L X di/dt
∫e dt = L X ∫di ;
i = 1/L X ∫e dt
if e = sin wt , i = 1/L X ∫sin(wt) = -1/ωL X cos(wt) if i didnt miss a sign someplace

and cosine is just sine shifted ninety degrees ..
Draw it out ?

 

[CosmicC]

First you have to define voltage as some math function.

Your question infers sine function but doesn't say that's what it is. Sine is a mathematical oddity in that its derivative and integral have its same shape .
so we use them almost interchangeably

Once you realize that it's trivial So you're aware e = L X di/dt
∫e dt = L X ∫di ;
i = 1/L X ∫e dt
if e = sin wt , i = 1/L X ∫sin(wt) = -1/ωL X cos(wt) if i didnt miss a sign someplace

and cosine is just sine shifted ninety degrees ..
Draw it out ?

Yes Now i get it. And even both the curves has difference of ninety degrees. Thanks a lot.

 

[CosmicC]

So we can mark this one as solved ? Or is there a remaining question ?

Solved. Thanks a lot. :)