Electric Circuit: i=i_{o}(1-e^{-t/\tau}) - Is (1) or (2) True?

[tuananh3ap]

i have one question
we know the electric cirlcuit RL
i=i_{o}(1-e^{-t/\tau})
with \tau=L/R
if R=0 then i is very big (1)
but
if \tau is very big then i increases slowly (2)
is (1) true or is (2) true

 

[jablonsky27]

hi,
the \tau in the formula doesn't determine the steady-state value of current. it only decides how soon the current will reach its steady state value.
if R=0 then the current will be big in only a purely resistive circuit. What you have is a resistive and inductive circuit. so if R=0, then the circuit becomes purely inductive. in a purely inductive circuit, the current reaches its steady state value infinitely fast.

 

[Redbelly98]

For R=0 in a series RL circuit, the current is a ramp function with constant slope.

So in a sense, both statements are true:

1. The "final current" is infinite.
2. It takes an infinitely long time to reach the final current.

 

[tuananh3ap]

thanks
i think your answer like my answer
but what is the end

 

[Redbelly98]

The current is a linear function of time, with a constant slope. There is no final value of current, it just keeps increasing forever.

 

[tuananh3ap]

thank you very much

 

[Domenicaccio]

For R=0 in a series RL circuit, the current is a ramp function with constant slope.

So in a sense, both statements are true:

1. The "final current" is infinite.
2. It takes an infinitely long time to reach the final current.

Sounds like a purely mathematical result rather than what physically happens tho...

If you really take a battery and connect its poles via an inductor, what happens? Does the inductor "extracts" all the energy from the battery, increasing the "stored" current linearly until the battery runs out, then the current remains circulating in the circuit (minor losses notwithstanding)? At least, the battery cannot be infinite, so the "i" has to stop at one point. But how does it take infinitely long to do so?

 

[Redbelly98]

You're right, I was giving a theoretical result for an ideal voltage supply with no current limit.

Once the current nears the maximum the supply can deliver, forget the calculation. You'll essentially get the supply's short-circuit current. If it's a battery, then even that will change as the battery runs down.