what would be the value of Vc(zero minus/plus) ie Vc(0-)=Vc(0+) . Is it [3*48/(48+100) =]97.30 or [3*48=]144 . I cant understand why the answer 97.30 is given in the book Hayt and Kimmerly 8th edison-- the source is a current source- not a voltage source, how the voltage division formula is applied here. If for above 97.30 is true then how come for the circuit shown below answer for Vc(0-)=48: According to the author Vc(0-) needs to be [3*48/(48+24)=] 2 rather than 48 [ =3*{24/(24+48)}*48] right?
This looks like homework. Even if it isn't homework and you are doing it completely on your own, you will probably get more attention and assistance over there.
If you track your units, you will see that the following: 3*48/(48+100) = 97.30 cannot be correct since it is (3A)(48Ω)/(48Ω+100Ω) which will result in an answer that has units of current, not units of voltage. To find the value of Vc(0-), analyze the circuit as it exists prior to t=0. Since the current source will have been at 3A for a very long time and nothing else is changing, what will the capacitor and inductor behave like at that time? So what does the effective circuit look like? In addition, you need to always ask if the answer makes sense. Does it seem reasonable that that expression can yield a value that is almost 100? You basically have 3 * (50/ (50 + 100) ) = 3 * (50/150) = 3 * (1/3) = 1 So you know the answer it will yield will be close to 1, not 100. Always, always, ALWAYS track your units! Always, always, ALWAYS ask if the answer makes sense!
Branches in parallel have the same voltage. Which means that if you find voltage across 24 Ohm resistor, you find voltage across the capacitor.
Thanks for the advice WBahn; we need to keep an eye on the "units". Just troubling you a bit more, can u plz let me know whether the 3 calculations [shown below] are right or wrong for the given circuit So (current through resistor) --> Ir(0-) = 3*48/(48+100) = 0.973 A and Vc(0-) = 3*48= 144 V and (current through inductor) -->Il(0-) = 3*100/(48+100) = 2.027 A And......plz have a look at this note: At t=0- ; the INDUCTOR acts as short circuit and CAPACITOR as open circuit to DC At t=0+ ; INDUCTOR replaced with current source[ Il(0+)= Il(0-) ] and CAPACITOR replaced with voltage source[ Vc(0+) = Vc(0-) ] and have to remove 48ohm and also the 3A current source.....WHY we need to remove this 3A curr source at t=0+ ?
As to your last question, look at the equation that governs the current of the current source. I = 3u(-t) A What is the value of u(-t) for t>0?
Let's at least give the TS an opportunity to apply what they know to answer such questions, that way if they can't answer we can identify that and resolve it.