Why does current decrease to a steady state value in a series RL circuit?

[FOIWATER]

I am under the impression that if you have an inductor and a resistor connected in series with a DC supply, that at the instant the switch is closed, the minimum current that circuit will flow, is flowing at this time?

kirchoff's law for the circuit... Vsource - Ldi/dt - Ri = 0
therefore, Vsource (as a function of time) = Ldi/dt + Ri.
divide both sides by L, V/L = di/dt + (R/L)i
this is a linear differential equation... find integrating factor to be e^(R/L)t
using power rule for derivatives once multiplying all terms by the integrating factor gives the well known formula - i(t) = Eo/R + Io*e^-(R/L)t --> where Eo and Io are applied voltage and initial current respectively.

This would mean, that the maximum current flows once the switch is closed? and it DROPS to steady state valuer as the transient goes to zero as t-->infinity. Intuitively, I would think instead that the current INCREASES to a steady state value as time approaches infinity?

Am I missing something.

 

[FOIWATER]

in fact the graph in this link suggests what I said does it not?

http://macao.communications.museum/...loor/moreinfo/2_3_6_ResistanceInductance.htmlAhhhhhhh... will Io (initial current) be negative?...

 

[phinds]

Am I missing something.

Think of it this way, for DC circuits

capacitors resist voltage --- you have to force voltage onto them, so that means that if you have a V-R-C circuit, all in series, the cap resists voltage and thus has to allow large current at first, but over time, it charges up and then has no current flowing through it (acts like an open circuit).

inductors resist current --- you have to force current through them, so that mean s that if you have a V-R-L circuit, all in series, the inductor resists current and thus has to have a very high effective resistance at first, thus takes all the volts that the circuit wants to put on it. Over time, the inductor allows current through (analogous to a capacitor charging up) and goes to a short circuit with no voltage across it.

 

[milesyoung]

You have a sign error. Have a look here:

http://www.electronics-tutorials.ws/inductor/LR-circuits.html

You subtract an exponential decay from your steady state value.

 

[FOIWATER]

yes but that is not what the equation suggests, unless the Io term is negative, because you are ADDING the transient term to the initial current. work it out on a calculator.. for small values of t, e^-(R/L)t is as almost one. meaning if Io is positive it gets ADDED to the steady state term. meaning MORE current at LOWER times. Which makes no sense to me and based on what you said it also wouldn't agree with you. The current as the circuit is closed should be LOWER since the inductor OPPOSES the current change. The equation doesn't relate to what I see happening... unless the Io term is actually negative... which is what happened when I set current and time to zero. I got a negative initial current... meaning I would now SUBTRACT the transient term, rather than add it.

Surely that must be it.

(Any question specific feedback appreciated).

 

[FOIWATER]

You have a sign error. Have a look here:

http://www.electronics-tutorials.ws/inductor/LR-circuits.html

You subtract an exponential decay from your steady state value.

YES - it is sort of hidden in my equation (my math is correct) but the value of initial current once we solve for for it, is actually a negative value -!

THANKS

 

[Ratch]

FOIWATER,

...but the value of initial current once we solve for for it, is actually a negative value -!

No, it's not. See attachment.

Ratch

 

[I_am_learning]

The solution for RL circuit with initial current will be like this (Yours is wrong)
I(t) = V/R (1 - e^(-(R/L)t) ) + I0e^(-R/L)t

Depending upon the direction of initial current (i.e sign of I0) the graph will look like this
http://www.wolframalpha.com/input/?i=Plot+5(1-e^(-t))-3*e^(-t)+from+t=0+to+5,+y+=+-5+to+6
(Taking V/R = 5, R/L = 1 and I0 = -3)
Or like this
http://www.wolframalpha.com/input/?i=Plot+5(1-e^(-t))+3*e^(-t)+from+t=0+to+5,+y+=+-5+to+6
(Taking V/R = 5, R/L = 1 and I0 = +3)

 

[Ratch]

You are learning,

The problem does not specify an initial current.

It would be nice if you specified to whom you are addressing your post.

Ratch

 

[AlephZero]

The problem does not specify an initial current.

It doen't need to be specified. The current for time < 0 is clearly 0. Because of the "L di/dt" term in the equations, the current cam't physically make a step change (because the the back EMF generated by the inductor would be then be "infinte"). So the current at time 0 is also 0.

When the switch is closed, the current will start to increase from zero so the back EMF equals the applied voltage, in other words at V/L amps/sec. The rate of increase will then reduce exponentially and the current will never exceed V/R amps.

 

[Ratch]

AlephZero

It doen't need to be specified. The current for time < 0 is clearly 0. Because of the "L di/dt" term in the equations, the current cam't physically make a step change (because the the back EMF generated by the inductor would be then be "infinte"). So the current at time 0 is also 0.

The current at time t<0 is assumed to be zero because the problem does not state otherwise. There could be a finite initial current in the coil at t<0, which would change the equation.

When the switch is closed, the current will start to increase from zero so the back EMF equals the applied voltage, in other words at V/L amps/sec. The rate of increase will then reduce exponentially and the current will never exceed V/R amps.

Correct, and doesn't my derived equation show that?

Ratch

 

[I_am_learning]

You are learning,

The problem does not specify an initial current.

It would be nice if you specified to whom you are addressing your post.

Ratch

I was replying to the OP.

i(t) = Eo/R + Io*e^-(R/L)t --> where Eo and Io are applied voltage and initial current respectively.

He assumed a solution involving the term Io (initial current), so I thought he was interested in solutions having initial current.
Of course, in the usual series arrangement, the inductor can't have initial current (before switch is closed) becasue there is no path to flow; But we can assume that somehow (clever ckts) can manage initial inductor current when the switch is closed.

Either way the nature of solution is same. If initial current were 0, the source will push the current to Eo/R through 0. If it were -Io, then it would need to push it from -Io to Eo/R and If it were Io, it would need to push it through only Io to Eo/R. All of these pushing occurs exponentially with a time constant of L/R.
I think everybody agrees here.

 

[psparky]

I am under the impression that if you have an inductor and a resistor connected in series with a DC supply, that at the instant the switch is closed, the minimum current that circuit will flow, is flowing at this time?

This is correct.

For RL circuit, any math I have ever done shows minimum current at t=0+ and nearly full current after 5 time constants.

As you know, your bandwidth and break frequency can be derived from the time constant as well.

 

[Emilyjoint]

It is pretty meaningless to say that capacitors 'resist' voltage... resistors resist voltage, also voltages cannot be forced 'on' anything, voltages appear ACROSS components. (current flowsTHROUGH...)
Also inductors do not 'resist' current, you should say (if anything) that inductors resist CHANGING current... this means that sometimes (switch off !) they ASSIST current...or try to...by the effects of induction.
More care should be used to use correct physics language

 

[Ratch]

Emilyjoint,

...More care should be used to use correct physics language

That sure is true. For instance:

The phrase "current flow" is redundant and ridiculous. Since current is charge flow, current flow means literally "charge flow flow". One should say instead that current exists.

Charge does not flow through the dielectric of capacitors. It accumulates on one plate and depletes on the opposite plate. Therefore the net charge of a capacitor is zero, whether it has zero volts across it or 100 volts across it. So, one should not say a capacitor is "charged", say instead that it is "energized". Caps do not store net charge, they store energy in an electrostatic field.

Voltage is the energy density of the charge, measured in joules/coulomb. Because density cannot be stored, neither can voltage be stored. It is instead accumulated.

Ratch