# circuit analysis

Discussion in 'Homework Help' started by fran1942, May 5, 2011.

1. ### fran1942 Thread Starter Member

Jul 26, 2010
58
0
Hello, I have the attached circuit which I am trying to solve.
I 'think' I have all the currents correct, however I am struggling to make the voltage drops over the resistors in the right hand loop(2) equate to 0 as per Kirchoff's law of voltages.
The two resistors on the left hand loop(1) drop the power source to 0, so that loop is fine. It is the big loop on the right that I am having trouble with.

If someone could please take a look and tell me where I am going wrong that would be much appreciated.

Thanks kindly.

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Last edited: May 5, 2011
2. ### StayatHomeElectronics Well-Known Member

Sep 25, 2008
864
40
R4 is not in the loop for the voltage calculation. And, the voltage drop over the final resistors should also give you a (-) sign.

12 - 4 - 2.67 - 5.33 = 0

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3. ### hgmjr Moderator

Jan 28, 2005
9,030
214
The reason that your voltage summation fails for the right-hand side of your circuit is that you are using voltage multiple time. Notice that if you had stopped at 12-4-8 = 0 you would have solved the first loop. The 2.66v and 5.333v figures associate with the rightmost loop only.

hgmjr

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4. ### miguel cool New Member

Mar 15, 2010
9
0
For lopp 1:
12 = I1 (9Ω + 1Ω) so I1 = 12/10Ω = 1.2Amp

but in the opposite direction that you draw I1

For the loop 2

R2 = 2Ω . . . . Req = 12Ω //6Ω gives 4Ω
so the total resistance in the right side is Rt = 6Ω = 2Ω+4Ω
and current I2 = 12V/6Ω= 2Amp that flow throug R2