Why current rises linearly in an inductor when constant voltage is applied?

[matbob]

Dear all,

Can someone tell me why the current through an inductor rises linearly when a constant voltage is applied across an inductor. Please don't explain in terms of equations because I know V = L * dI/dt and so I is the integral of voltage applied over time and therefore I = ( V * t ) / L. I would appreciate if someone can tell me the exact mechanism, in terms of the magnetic flux/field, self-induced voltage etc.

Regards,
matbob

 

[jeffrey samuel]

inductor in dc supply is short circuited and so it behaves like a mere resistor

 

[dick_freebird]

V=L*dI/dt, V and L are constant, hence dI/dt is constant.

 

[godfreyl]

1. The magnetic field strength is proportional to the current.
2. The induced voltage is proportional to the rate of change of the magnetic field strength.
3. Therefore the induced voltage is proportional to the rate of change of the current.

I have to admit I don't really understand the reason for (2) above.

Please don't explain in terms of equations...

Heh heh! You're assuming people will read your whole post before replying. Looks like Jeffrey didn't even read the heading.
:grin:

 

[Ratch]

matbob,

Can someone tell me why the current through an inductor rises linearly when a constant voltage is applied across an inductor.

No one can. You would do well to look at the equation for current through an inductor when a constant voltage is applied. It is I = (E/R)(1-exp(-t/(L/R))). As you can see, current through a inductor has an exponential relationship, not a linear one.

Ratch

 

[jeffrey samuel]

Heh heh! You're assuming people will read your whole post before replying. Looks like Jeffrey didn't even read the heading.
:grin:

from experience i learnt to become a better person not to just read the heading

and really true i did not go through the heading of the post fully

your explanation is correct as i find the fact is in general all memory devices behave linearly in constant supply voltage

memory devices uses the rate of variations produced (frequency) of the signal input and respond diversely and show their corresponding characteristic curve

if the rate of change of ip is 0 (as in the case of constant sources) they behave linearly (inductor) or as an open circuit (capacitor) as there is no point where these devices can discharge their memories (voltage or current) to the external circuit

 

[godfreyl]

Please stop writing nonsense. It does not help anybody. Inductors and capacitors are not memory devices.

memory devices uses the rate of variations produced (frequency) of the signal input and respond diversely and show their corresponding characteristic curve

That sentence doesn't even make sense. Maybe your translation software is broken.

 

[jeffrey samuel]

you are forgetting the fact that the cap's op depends on the ip voltage given which is stored inside it as charges in the plates

that is what i meant

you are forgetting something else if cap has no memory then if supply is cut there is no possibility of regaining the charge stored in it

cap's and inductors use it i never meant other mem devices like ROM AND STUFF

am i now a bit more clear