Calculating current through an IR circuit

  • #1

Main Question or Discussion Point

Hi, if I have a DC indoctor-resistor circuit with and inductance (L) of 700μH and a resistance (R) of 0.8Ω, I realise that I can use:

IL = (Vsource/R) * (1-e-t*(R/L))

to calculate the current through the inductor if I have a constant voltage source, but would it be possible to calculate how the current build (and also decay) might look in this or a similar method if my source voltage looked more like:

http://www.dransfield.talktalk.net/VoltageTrace.jpg [Broken]

So far I can't work out how to make this work with a changing voltage, as the equation appears to assume that you always start at 0

Also, on the decay side, the equation:

IL = (Vsource/R) * (e-t*(R/L))

Seems to assume that you always start at current I=V/R, which is not always necessarily true

I hope I've explained myself well enough, and thanks for any help in advance!
 
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Answers and Replies

  • #2
tiny-tim
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Hi LADransfield! Welcome to PF! :smile:

LdI/dt + RI = V(t)

so use an integrating factor to simplify the LHS into a single integral, and then use area-under-a-graph :wink:

(btw, it's not called an IR circuit, it's an RL circuit)
 
  • #3
Ah, thanks very much, I know it was something similar, but got the wrong letter!

In this case, which "I" would I use in RI? Would this also be the induced current, or just the dI/dt?
 
  • #4
tiny-tim
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In this case, which "I" would I use in RI? Would this also be the induced current, or just the dI/dt?
uhh? :confused:

I is I

dI/dt is d/dt of I

(and the current through R is the same as the current through L … where else would it go?)
 
  • #5
I was under the impression (from here and a few other places) that the total current was I = V/R , but there were different amounts in the inductor and resistor as the current has to build over time in the inductor (iL), but the current in the resistor (iR) was I-iL

I am expecting the current to rise sharply due to the high voltage, then remain a little more constant through the lower (hold) voltage

Simulation and testing shows that this happens, though creating a more one-dimensional approach is proving a little more tricky
 
  • #6
tiny-tim
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Hi LADransfield! :smile:
I was under the impression (from here and a few other places) that the total current was I = V/R , but there were different amounts in the inductor and resistor as the current has to build over time in the inductor (iL), but the current in the resistor (iR) was I-iL
From your link …
"once the Storage Phase has finished, the Current, iL, that flows through it is stable, iL = V / R, no Self Induced e.m.f. is produced and vL is zero. The Inductor acts like an ordinary connecting wire, its Resistance is zero."​
… iL only = V/R when steady-state has been reached

until then, the total voltage = the voltage drop across the resistor (IR) plus the "induced" LdI/dt across the inductor.

It's the same I across both, for the same reason you have the same current through a water pipe that goes through various attachments …

water in = water out, and electric current in = electric current out. :wink:
 
  • #7
In that case, how do I calculate the induced current at any time-step using a changing voltage input?

The equation you specified doesn't appear to calculate the induced current, which is what I'm after?
 
  • #8
tiny-tim
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In that case, how do I calculate the induced current …
the current I is the current through the inductor

it's "induced voltage" (LdI/dt) across an inductor, not "induced current"

(eg if you look up "induced current" in wikipedia, you'll be redirected to a page which doesn't mention "induced current")

an inductor is a voltage source (emf source), not a current source :wink:
 

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