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Finding the current on each resistor

  • Thread starter kliker
  • Start date
  • #1
104
0

Homework Statement


[PLAIN]http://img297.imageshack.us/img297/7053/96217079.jpg [Broken]

given this circuit find the current on R1,R2 and R2


Homework Equations


I = V/R


The Attempt at a Solution



i applied kirchhof's law and I have this:

36 - 5*I - I1*4 = 0
36 - 5*I + 20 - 2*I2 - 14 = 0
20 - 2*I2 - 14 + I1*4 = 0

and i find I1 = 66/19 which is wrong
the book says
I1 = 1.10 A
I2 = 5.21 A
I = 6.32 A

what am i doing wrong?
 
Last edited by a moderator:

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
Hi kliker! :smile:

(try using the X2 tag just above the Reply box :wink:)
I applied kirchhof's law and I have this:

36 - 5*I - I1*4 = 0
36 - 5*I + 20 - 2*I2 - 14 = 0
20 - 2*I2 - 14 + I1*4 = 0
The circuit only has two independent loops, so you can only get two independent equations from it.

The third Kirchhoff equation you need is for one of the junctions. :smile:
 
  • #3
104
0
Hi kliker! :smile:

(try using the X2 tag just above the Reply box :wink:)


The circuit only has two independent loops, so you can only get two independent equations from it.

The third Kirchhoff equation you need is for one of the junctions. :smile:
hi, thanks for the answer

if we say I = I1 + I2 and take the first two equations

then we will have

36 -5*I1 - 5*I2 - 4*I1 = 0
36 - 5*I1 -5*I2 +20 - 2*I2 - 14 = 0

36 - 5*I2 - 9*I1 = 0
36 - 5*I1 - 7*I2 + 6 = 0


36 - 5*I2 - 9*I1 = 0
42 - 5*I1 - 7*I2 = 0

then by solving the equations I found the desired results

thanks a lot

Ps: what's the x2 tag? I can't find it
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
251
Hi kliker! :smile:

(just got up :zzz: …)
…then by solving the equations I found the desired results
:biggrin: Woohoo! :biggrin:
Ps: what's the x2 tag? I can't find it
Press the "QUOTE" button or the "Advanced Reply" button, and you get to the Reply to Thread page, with loads of useful tags to play with, above the box where you enter your text. :wink:
 

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