i tried that, atleast i think i did. it is for an exam i am taking. I have the correct answer but cant remember how i got it. Imagine a square circuit where R1 R2 and R5 are on the outside perimeter in series. Then R3 is branched in parallel between R1 and R2 and R4 is branched between R2 and R5. R1=25 ohms R2=25 ohms R3=50 ohms R4=5o ohms R5=50 ohms. Total volts is 50. Total current is 1 amp. Those are all given on exam. I need to find the current flow thru R3. Got any ideas?
one way is
go to post reply button
u'll see a attaching clip button after the smiley icon.
if not
u'll see manage attachment button when u scroll down a little low.
however i think i m getting the picture here, but still post it
also where is the voltage applied?
i hope this worked. the last one i tried typing into the forum reply box but it shifted the branches over to the left instead of between R! and R2. I hope this works.
yeah it will work thanks,
there are lots and lots of ways to do it.
number one simple analysis:
look at the last two resistances.
they are in parallel, hence R4||R5
will give u 25 ohms (u know how to calculate it ,dont u?)
now the two 25 ohms
R2 and R4||R5 are now in series.(replace parallel combn of R4 and R5 by eq 25 ohms)
you get 25+25=50 ohms,
again R3 and R(equivalent=50) are ||
again they give equivalent of 25 ohms,
R1 and Req are in series thus we have 25+25=50 ohms,
apply ohms law i=50/50 = 1 amp
now again use original cicuit;
since now current is 1 amp;
there will be 25*1 v drop across r1;
now remaining 25 volts is available across the whole remaining circuit,
since r3 is parallel to remaining circuit 50-25*1 =25 volts is available across
r3;voltage across parallel elements is same.
thus current thru r3 is 25/50 =0.5 amp
another method:
after getting current = 1 amp;
go back to step two where we had r3 parallel to a 25+25 =50 ohms combn;
now use the current divider rule it is;
for two parallel paths the current in one path is given by:
I1= I(original)*{res other path/(res req path + res other path)}
here i original is 1 amp which reaches the parallel combination of r3 and remaining circuit;
req resis is r3 and resis in other path is R4||R5 + r2
another method
this method is universal and can solve virtually anything:
use kvl (in case u want to know how to use it then try reading any book or
article in this site-if u still have problem post it)
no problem:
remember whenu have any doubt try your best;
u'll learn more.
also try solving in more than way this gives better understanding of circuits and their laws.
good luck
i was taking R3 R4 and R5 and using them as being in parallel \with the reciprical method. I have been working on this for about 5 hrs knowing the answer was .5 amps but could not get there. once i used the R4xR5 over R4 +R5 formula to get req then R3xReq over R3 + REQ i had it figured out. Thanks again. REally.
that mistake is possible,
remember two elements r in parallel when their ends are joint at 2 common points such that there is no other element in between.