Can you give me some help,suggestions about these circuits ?

recca02

Joined Apr 2, 2007
1,212
confusions,confusions..
one small question why is i=3a given i dont see a current source?
and there are more resistances than given
shud i ignore them as u mentioned,
also with all those components neglected (i think we are dealing with dc sources here hence l gets shorted and c gets open circuited)the circuit is same as u drew ,did ur friends get the correct answer and the same circuit?
also which one is r1 and r2?, a small error here and there can lead to great difference in values.
shud i take the ones in ur simplified circuits as the correct resistance values?
btw r2 seems to be r7 to me.
is that big i coming into the circuit or what ?(i did not take that into account last time while writing equations, i noticed it now)
 
I=3a is given because we have a source - one up and one on the left.You should ignore those not needed.r1 is between 1st loop and 3rd loop (upper left and lower left triangle).We got the correct answer on another circuit with the same datas and the "teacher" has given one data for all the circuits so it mus'nt be a problem.Yes you should take the ones in the simplified picture.It's big I.Does it have to do something with the equations of the Kirchhoff's law...
 

recca02

Joined Apr 2, 2007
1,212
yeah the big I is a big problem here due to it all my previous equations i believe are going to change, right now i m not able to recall what shud be done when there is a current source acting with a voltage source, i'l look it up an see if i can remember it,after that solving the circuit shud not be a difficulty(damn i miss my old book) i will edit this post later. check back after sometime.
give me the value that voltmeter measures;
i m going to try some tricks and see if i can land up with a correct answer (like i said i cant recall what effect a current source in series has with voltmeter - ideally when no resistance is present the voltage source is killed, but i dont think that wud work here)

for the second question are the sources again dc ?
do we have to use the mutual coupling in this case since polarities are marked?
 

recca02

Joined Apr 2, 2007
1,212
ok,
since i could not find the book (my fav book :( )
i tried it the old superposition way,
i really feel sorry for posting this so late;
i did some rough calculations can u ppl post the voltmeter answer i got it as 29.5 v of course chances of error are there but i'll let u know how i approached it;
u can use the equations i posted before but u'll have to separately take in to account the effect of the current source, remove the voltage sources and using current divider rule or ckt simplification find current thru r3 and r6,
add up voltages(algebraically) found by above to calculations and u shall have the correct answer.
i could have posted the answer for second ckt but u ppl didnt reply.
if some one can come up with a better method pls post.
 
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