YaEET: Unit Step Function

[Ricemarine]

Look at 7.20, Fig. 7-33

I'm looking at the problem for the unit step function. Now the way I thought of solving for i(t<0) is that you can KVL the 50u(-t), the inductor, and the two 10 ohms resistors. Which gets me the answer of 2.5A as stated for that problem...

Now one issue that was raised up when coming to that conclusion was is that 10 ohm resistor being bypassed around the 5 u(t) current source? For if there was no unit source there, is it saying there is 0 A in the current or is there still some current through that branch?


Thanks...

btw, this section seems to be much much easier to do than the last homework assignment I had. Putting it into smaller problems... works very well .

 

[Minjin]

For t<0, the current source is off. That means current = zero and it is equivalent to an open. Your circuit is then just the two resistors and inductor, all in series.

Remember:

voltage source turned off => voltage across = 0 => acts as a short
current source turned off => current in branch = 0 => acts as an open

 

[Ricemarine]

Originally posted by: Minjin
For t<0, the current source is off. That means current = zero and it is equivalent to an open. Your circuit is then just the two resistors and inductor, all in series.

Remember:

voltage source turned off => voltage across = 0 => acts as a short
current source turned off => current in branch = 0 => acts as an open

Okay, that makes sense... but now that raises another question... There's another problem in my problem set which I'm not sure if that applies too (if you don't mind).

---3 ohm-----6 ohm---
|-----------|------------|
^10u(t-1)S (2H) 10u(t) ^
|________|________|

S = squiggly inductor thingy.
^ = plus sign is up


So if what you say is right. So initially... for this problem too, with the 10u(t) volt source., There is current through both the 3 ohm and the 6 ohm, and since as you said, the 10u(t-1) is a short, then Req = 2 ohms (3*6/9). So then the equivalent current through the inductor would be 5 amps?... (10 V / 2)

 

[Minjin]

I have no idea what your drawing shows. Throw it up on paint.

Or something like this:

http://www.flashpaint.com/home.php