# Need Heeeeeelp Plz

#### one way

Joined May 27, 2007
7
I need some help figuring current flow across a resistor in series-parallel circuits.

#### beenthere

Joined Apr 20, 2004
15,819
Determine the potential across the resistor, use I = E/R to figure the current through it (works in series and in parallel).

#### one way

Joined May 27, 2007
7
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?

#### recca02

Joined Apr 2, 2007
1,214
i m not getting the dia at all (must be my poor skill with language)
plz post rough dia (a super bad pic made in paint will also do)

#### one way

Joined May 27, 2007
7
---------R1 25 ohms-------o-----------------R2 25 ohms----o------------
! ! !
! ! !
R3 R4 R5
50 ohms 50 ohms 50
! ! !
! ! !
! ! !
-------------------------------------------------------------------------

total volts is 50 total amps is 1. What is current draw across R3? and how did you come up with that? Thanks alot for your help.

#### one way

Joined May 27, 2007
7
i have a really ruff sketch in paint how do i add to thread?

#### recca02

Joined Apr 2, 2007
1,214
one way is
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?

#### one way

Joined May 27, 2007
7
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.

#### Attachments

• 396.8 KB Views: 22

#### recca02

Joined Apr 2, 2007
1,214
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)

there are still other ways to solve it.

attachment may help u understand better

#### Attachments

• 29.6 KB Views: 19

#### one way

Joined May 27, 2007
7
Thanks alot. I was getting really frustrated. I really appreciate it.

#### recca02

Joined Apr 2, 2007
1,214
no problem:
remember whenu have any doubt try your best;
also try solving in more than way this gives better understanding of circuits and their laws.
good luck

#### one way

Joined May 27, 2007
7
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.

#### recca02

Joined Apr 2, 2007
1,214
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.