Driven RL circuit, why at infinity is this the current 2, and yet at 0+, its 0?

In summary, a driven RL circuit will have a current of 2mA at time infinity and 0 at time 0+. This is because after a long period of time, the inductor will act as an ideal wire, allowing all current to flow through it instead of the resistor. At time 0+, the inductor will act as an infinite resistance, causing all the current to flow through the resistor. This also applies to sourceless RC/RL circuits, where at t=0+ the inductor will act as an infinite resistance and the capacitor will act like an ideal wire if it was initially uncharged. If it was initially charged, it will act like a battery with a potential difference determined by the initial charge and capacitance
  • #1
mr_coffee
1,629
1
Hello everyone I'm confused...i was just doing RL and RC circuits without a driven source, they made sense after doing them a few 100 times. Now I'm at Driven RL circuits. My professor said:
iL(infinity) = 2mA
iL(0+) = 0.

From my understanding, i thought at infinity, the switch has been closed for a long period of time. If the switch is closed for a long period of time that means there would be no current flowing through the inductor, because you have to have changing current. So why isn't THe current of the inductor 0 at time infinity?>

At time 0+ that means the switch has just been closed, meaning your going to have a change in current. So wouldn't that be when the current through the inductor is 2mA?

Am i interpretting the unit step wrong?
2u(t) ?
this formula doesn't ask for time 0-, that means before the switch is closed, now at that point i can see how she would get iL(0-) = 0, becuase the current is totally disconnected!

Any help on explaining this would be great! :)
Also by doing it her way, she got the right answer in the back of the book.

http://suprfile.com/src/1/bx0wa2/lastscan.jpg

THen she found VL, the voltage on the inductor and got the opposite, she wrote:
VL(0) = 2 mA* 3k = 6v
vL(infinity) = 0;
VL = 6e^(-t/5E-6);
x = 15mH/3k = 5E-6;

She also wrote this which i think explains why,

Current in inductor and Voltage on a capacitor can NOT change instantly
Voltage on inductor and current in a capacitaor can change indstantly


So is this why she has iL(0+) = 0? it takes time for the 2 mA to go through the inductor?
 
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  • #2
mr_coffee said:
Hello everyone I'm confused...i was just doing RL and RC circuits without a driven source, they made sense after doing them a few 100 times. Now I'm at Driven RL circuits. My professor said:
iL(infinity) = 2mA
iL(0+) = 0.

From my understanding, i thought at infinity, the switch has been closed for a long period of time. If the switch is closed for a long period of time that means there would be no current flowing through the inductor, because you have to have changing current. So why isn't THe current of the inductor 0 at time infinity?>
No, after the switch has been closed for a very long time, an inductor will act as it it was an ideal wire! So in that case, all the current will go through the inductor and none through the resistor (since the current has the choice of flowing through a branch offering no resistance at all, it will go that way). At t=0+, the change of current is huge (theoretically infinite) so the inductor offers an inifinite resistance, so all the current will flow through the resistor.
 
  • #3
Thanks again for clearing up that! :biggrin:
What if i replaced that inductor with a capacitor? With sourcless circuits I treated the capaictor as if it was an open circuit if no current has been flowing or if its been flowing for along time.

nrqed, is this the case only for source driven RL/RC circuits? or does it also apply to sourceless RC/RL?
 
  • #4
mr_coffee said:
Thanks again for clearing up that! :biggrin:
What if i replaced that inductor with a capacitor? With sourcless circuits I treated the capaictor as if it was an open circuit if no current has been flowing or if its been flowing for along time.

nrqed, is this the case only for source driven RL/RC circuits? or does it also apply to sourceless RC/RL?

In general, you have the following (for DC circuits):

If a circuit has been in a state for a very long time, inductors will act like ideal wires (R=0) and capacitors will act like infinite resistances (no current flows through their branch. They have a potential difference across them and Q= C V).

If change is made (by closing, opening a switch) then an inductor will act (at t=0+) like an infinite resistance. For capacitors, it depends if they were initially charged or uncharged. If they were initially uncharged, at t=0+ they will act like ideal wires (R=0) (Of course, this is just an analogy..no current flows through them, but for the purpose of calculating solving the circuit they can be treated like ideal wires at t=0+).
If they were initially charged, they can be treated (still, at t=0+) like batteries with a potential difference V=Q_0/C where Q_0 is the inital charge.

Patrick
 
  • #5
Thanks a ton Patrick, excellent summary!
 

1. What is a driven RL circuit?

A driven RL circuit is a type of electrical circuit that contains a resistor (R) and an inductor (L) connected in series, with an external power source (driving source) supplying an alternating current (AC). This circuit is commonly used in electronic devices such as radios and televisions.

2. Why does the current approach 2 at infinity in a driven RL circuit?

At infinity, the inductor in a driven RL circuit behaves like a wire, allowing the current to flow freely without any opposition. This results in a maximum current of 2, which is determined by the voltage and resistance in the circuit according to Ohm's Law (I = V/R).

3. Why is the current 0 at 0+ in a driven RL circuit?

At 0+, the inductor in a driven RL circuit behaves like an open circuit, meaning that it offers infinite resistance to the flow of current. This prevents any current from passing through the circuit, resulting in a current of 0. This is known as the transient state, as the inductor needs time to build up its magnetic field and allow the current to flow.

4. How does the inductor in a driven RL circuit affect the current?

The inductor in a driven RL circuit creates opposition to the flow of current due to its property of self-inductance. When the external power source is turned on, the inductor initially resists the change in current, resulting in a transient state where the current is 0 at 0+. As the inductor builds up its magnetic field, it allows the current to flow, eventually reaching a maximum value of 2 at infinity.

5. Can the current in a driven RL circuit ever be higher than 2?

No, the current in a driven RL circuit is limited to a maximum value of 2, determined by the voltage and resistance in the circuit. This is a fundamental property of the circuit and cannot be exceeded, even with different values of voltage or resistance.

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