- #1
mr_coffee
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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?
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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