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feezul_93 Group TitleBest ResponseYou've already chosen the best response.1
How to find the thevenin for cd terminal ?
 one year ago

feezul_93 Group TitleBest ResponseYou've already chosen the best response.1
I mean, thevenin voltage :)
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
dw:1356752136345:dw open the current source and short the voltage source you will get Rth
 one year ago

feezul_93 Group TitleBest ResponseYou've already chosen the best response.1
I did, but I didn't get the right answer :(
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
hold on let me solve
 one year ago

feezul_93 Group TitleBest ResponseYou've already chosen the best response.1
I got 2.667 – j4 Ω for Rth, n the answer for Vth is 9.614∠56.31° V
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
did you convert 2.667 j4 into polar form , you've got right answer
 one year ago

feezul_93 Group TitleBest ResponseYou've already chosen the best response.1
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 ?
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
i think you should be using superposition principle
 one year ago

feezul_93 Group TitleBest ResponseYou've already chosen the best response.1
ehh ehh, isn't it suppose to be supermesh ? D: there's current source within the loop
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
there will be nothing messy , give a try by that first eliminate current source and then eliminate voltage one
 one year ago

feezul_93 Group TitleBest ResponseYou've already chosen the best response.1
ouhh, okayy, but the concept Im using just now is right right ?
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
yep, its right i am sure you'll get your thevnin voltage by that
 one year ago

feezul_93 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
well, Vth, you can use voltage division rule
 one year ago

feezul_93 Group TitleBest ResponseYou've already chosen the best response.1
hold on, currently using superposition to solve it :)
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
dw:1356753391535:dw calculate voltage drop at point a and b
 one year ago

feezul_93 Group TitleBest ResponseYou've already chosen the best response.1
dw:1356753632924:dw
 one year ago

feezul_93 Group TitleBest ResponseYou've already chosen the best response.1
oh, some text missing ><
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

feezul_93 Group TitleBest ResponseYou've already chosen the best response.1
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 @.@
 one year ago

feezul_93 Group TitleBest ResponseYou've already chosen the best response.1
can I just ignore the terminal 1st, n solve it like a normal question ?
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
you dont need to go through all that stuff , just use superposition and you are done
 one year ago

feezul_93 Group TitleBest ResponseYou've already chosen the best response.1
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 ?
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

feezul_93 Group TitleBest ResponseYou've already chosen the best response.1
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 :)
 one year ago

feezul_93 Group TitleBest ResponseYou've already chosen the best response.1
D: u can guess the answer just by watching it ? awesome ! D:
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
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 :)
 one year ago

feezul_93 Group TitleBest ResponseYou've already chosen the best response.1
I don't get it, why we do we have to let V1  V2 ...
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
we are finding V1V2 because just take off the capacitor and you are supposed to find the terminal voltage
 one year ago

feezul_93 Group TitleBest ResponseYou've already chosen the best response.1
soo, deducting v2 by v1 is equal too ... V at terminal ? D: I don't see relation between it, can't even imagine it _"
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
dw:1356762940423:dw look here these are V1 and V2
 one year ago

feezul_93 Group TitleBest ResponseYou've already chosen the best response.1
yeap2, but then ... v1 isn't connected with v2 D:
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago
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