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Curry
 one year ago
Question regarding circuits.
Curry
 one year ago
Question regarding circuits.

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Curry
 one year ago
Best ResponseYou've already chosen the best response.2the answer is 4/3 + 27.7 cos(20t  56deg). I was able to get the second part of the answer using super position. However, i don't know where the 4/3 is coming from.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0what differential equation did you solve? and how do you tell the initial condition? if you don't know I'm still browsing the internet

Curry
 one year ago
Best ResponseYou've already chosen the best response.2well, using super position, I first turned off the 2v voltage source. and then solved for v(t). then i turned off the current source and then solved for v(t) again. Then added these 2 v(t). That gives the second part of the answer.

Curry
 one year ago
Best ResponseYou've already chosen the best response.2and I didn't use any differential equation, i used phasors.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0oh looking up that term :p

freckles
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure if @ganeshie8 knows about circuits but I will call him here just in case

Curry
 one year ago
Best ResponseYou've already chosen the best response.2so solving that for current gives 2v([(10+10j)+(205j)] x (205j) = V1.

Curry
 one year ago
Best ResponseYou've already chosen the best response.2Solving that gives V2. V1 + V2 gives the 27.7cos(20t  56). But i don't get where the 4/3 comes frmo.

Curry
 one year ago
Best ResponseYou've already chosen the best response.2@dan815 @satellite73 @ganeshie8 @perl

freckles
 one year ago
Best ResponseYou've already chosen the best response.0http://www.labsanywhere.net/circuit/lectures/lect10/lecture10.php this seems to be similar to what you are doing

freckles
 one year ago
Best ResponseYou've already chosen the best response.0except they solved for i

Curry
 one year ago
Best ResponseYou've already chosen the best response.2well that circuit is different because it's both AC. My problem is AC and DC. :(

Curry
 one year ago
Best ResponseYou've already chosen the best response.2well that circuit is different because it's both AC. My problem is AC and DC. :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0CURRY plz come back and help me

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.12 * 20/30 = 4/3 ie not sure what you are doing but the 2V source surely only goes through the resistors total impedance to AC is \(5.55 e^{56^0j}\) giving the right hand part of the desired solution.

Curry
 one year ago
Best ResponseYou've already chosen the best response.2Why doesn't the 2v go through the inductor and capacitor? I've done a lot of problems with DC circuits and Capacitors and inductors.

Curry
 one year ago
Best ResponseYou've already chosen the best response.2And calculating the second part of the equation requires both the AC and DC value. When doing it through super position ofc.

Curry
 one year ago
Best ResponseYou've already chosen the best response.2btw, what major are you Irish boy?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1direct current goes through a capacitor? in a dcC circuit you will get that initial flow as the cap charges up but that's it. it "will" go through the inductor but what is the inductor's resistance? so the dc to superimpose is flowing through a 10\( \Omega\) and 20\( \Omega\) resistor and you're measuring the drop on the latter. the AC is a RL/R/C all in parallel. btw, if he's around, this is "the" chap to tag on this type of stuff: @radar

Curry
 one year ago
Best ResponseYou've already chosen the best response.2So, when i have a DC and an AC, what should my appraoch be? I usually start with super position. I calculate the voltage drop from the DC and then the AC. and then i add them up. So for this problem, i did that and i got the correct answer for the latter of the equation. but the initial part i couldn't. So generally, is it just, ignore the AC, and short inductors. and open capacitors?

Curry
 one year ago
Best ResponseYou've already chosen the best response.2@IrishBoy123 Hey can you clarify whenever you can?

Curry
 one year ago
Best ResponseYou've already chosen the best response.2wait nvm. I got it. Thanks so much. you saved my life.
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