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I mean, thevenin voltage :)
|dw:1356752136345:dw| open the current source and short the voltage source you will get Rth
I did, but I didn't get the right answer :(
hold on let me solve
I got 2.667 – j4 Ω for Rth, n the answer for Vth is 9.614∠56.31° V
did you convert 2.667 -j4 into polar form , you've got right answer
i got the Rth, but Vth ... I get it wrong :( wondering where's my mistake, Im using mesh analysis to find current flowing through capacitor, n times the capacitor to get Vth ... am I wrong ?
i think you should be using superposition principle
ehh ehh, isn't it suppose to be supermesh ? D: there's current source within the loop
there will be nothing messy , give a try by that first eliminate current source and then eliminate voltage one
ouhh, okayy, but the concept Im using just now is right right ?
yep, its right i am sure you'll get your thevnin voltage by that
but I didn't T^T nevermind, i try to use superposition now, thanks :) btw, did u get the Vth ? cause I got the feeling that the answer given is wrong, hahaha
well, Vth, you can use voltage division rule
hold on, currently using superposition to solve it :)
|dw:1356753391535:dw| calculate voltage drop at point a and b
oh, some text missing ><
actually you are supposed to calculate thevenin voltage across the capacitor that is i guess your load here, so you have to make it open terminal to calculate the voltage drop at load
yeah yeahh, that's exactly what I did just now ... n base on your drawing ... we're using nodal analysis ? cause I prefer to find current through capacitor n simply times it, but I don't really know if my concept is right ... there are too many ways to solve this @.@
can I just ignore the terminal 1st, n solve it like a normal question ?
you dont need to go through all that stuff , just use superposition and you are done
T^T, i get the same answer like my previous method ... I guess the answer given is wrong ... but still ... I can't clarify it if Im the only 1 who get this answer ... did you get the same answer as given ?
actually i hate these complex calculation lol so i can write the method you can match it , that is voltage drop due to 20 V source and due to current source
i'll upload the solution and if you are getting same answer then i guess your book has got some mistake, i am not sure but i guess its 4 angle 56 V
hahahahaha, yeaappp, same goes to me, and I wonder who create this stuff, what kind of brain he had -___- but i guess i understand the flow of answering this kind of question, ouhh really ? thanks a lot :)
D: u can guess the answer just by watching it ? awesome ! D:
you are welcome :D and yes the answer is around what i said, but i have to check it again , only reason is i hate this complex calculation lol i'll have to use calculator , dont worry i'll do it :D :)
we are finding V1-V2 because just take off the capacitor and you are supposed to find the terminal voltage
soo, deducting v2 by v1 is equal too ... V at terminal ? D: I don't see relation between it, can't even imagine it -_-"
|dw:1356762940423:dw| look here these are V1 and V2
yeap2, but then ... v1 isn't connected with v2 D:
when we measure potential difference we measure the potential difference at two terminals, that is why we have eliminated capacitor it is not needed to be connected to measure the potential drop across those terminals
if like you will join that two points by wire there will be zero potential difference which means both the points at same potential or short circuit
just think of using multimeter for the measurement of potential difference across a resistor or any element , we place two terminals of multimeter at the two ends