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one last Circuit question......

S

Semidevil

Diamond Member
ok, given the circuit where you have a 4.8 ohm resistor, 4H inductor, and a .25F capacitor all in a series, connected to a 160 V source, how do you find the s-domain expression for I.

so in S domain
Resistor = 4.8ohm
inductor = 4 S
capacitor = 1/.25 S or 4 S.

I did a regular KVL loop equation to get 4.8I + 4SI +1/.25S(I) =160V.

I need to get I, so I take I out of the left side, to get I(4.8 + 4S +1/.25S) = 160

now, I get I by itself:

I = 160/(4.8 + 4S + 1/.25S).

now I"m lost. what do I do next?

the answr the book gives is:

I = 40/(s^2 +1.2s + 1).

how?
 
ok, I"m not sure if I left out some important material, but there is a switch right next to the Voltage source, saying

"the energy in the circuit shown is zero at the time when the switch is closed."

does this help?

 
That plays a huge role. Let me get back to you.

Edit: I thought that the problem sounded odd, as cap/ind problems are almost always time dependent. The reason is because at steady state (caps and inds charged), capacitors appear as open circuits and inductors appear as shorts.
 
Originally posted by: WhiteKnight
That doesn't do it, there is an s^2 term in the book answer.
Click to expand...

Whoops. Didn't notice that.

Anyway, WhiteKnight is right in that the switch is important. Your voltage source now has a unit step response, which has a Laplace transform of 1/s. so 4.8Is + 4Is + 4I/s = 160/s

Edit: Doh, beat to the punch 😛
 
Glad I remember something. I finally "finished" my Circuits book last quarter. Stupid book took me 4 classes to go through. Felt quite good when I finished it😎
 
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