Solving I3 Resistor Problem with Kirchhoff's Laws

[Loke]

Find the current in I₃ resistor by using kirchhoff's laws.

problems:
-when i try to use norton theorem to check for my answer and my answer is I_N=2.855A=I₃ but when i use kirchhoff's law i get I₃=0.645A.
-i don't know where is the problems...the current for norton theorem and kirchhoff's law should be the same right?

 

[vela]

You need to show your work. We can't tell where the problem is if all you do it post two numbers and ask why they aren't equal. That said, as long as you solve the problem correctly, you should get the same result regardless of which method you use.

By I₃, I assume you mean the current flowing through resistor R₃. If that's the case, neither of your answers appears to be correct, based on my calculations.

 

[Loke]

Find the current in I₃ resistor by using kirchhoff's laws.

problems:
-when i try to use norton theorem to check for my answer and my answer is I_N=2.855A=I₃ but when i use kirchhoff's law i get I₃=0.645A.
-i don't know where is the problems...the current for norton theorem and kirchhoff's law should be the same right?

attempt solution:
-norton theorem
I_N=V_S/(R₁+R₄)=2.86A

-kirchhoff's law
20-2I₁-5I₁-5I₂=0
5I₃+5I₃-5I₂=0
I₁=I₂+I₃

 

[vela]

What exactly is I₃ supposed to be? And why would you expect it to be equal to I_N?

 

[Loke]

Ohh... in my kirchhoff's law equation, my I₂ is flowing through resistor R₃ and i get I₂ equal to -1.29A not I₃=0.645A i wrote wrongly ...whereas in my norton theorem equation, my I₃ flowing through resistor R₃...which is IN=2.855A...the answer is not the same...please advice.

 

[Loke]

by the way, can you show me your solution? actually this is not my homework... i just want to know how to do...because i almost exam...i just simply find a question and do...or at least tell me your answer...i work it out myself...i really running out of time...no offence...thanks...

 

[vela]

Your set-up using Kirchoff's laws is correct. Solving those equations, I found I₁=60/31, I₂=40/31, and I₃=20/31.

With the Norton approach, if you redraw the circuit after the transformation, you have the current source and three resistors in parallel. Your mistake is thinking all the current from the source will flow through only one of those resistors.

 

[Loke]

as for norton theorem approach, i shorted R₃ and i get I_N=V₁₄/R₁₄=V₂₅/R₁₄=20/7=2.855A i still got the same answer...and how to find R_N when i open R₃ and shorted voltage source ...R₁+R₂+R₄+R₅=17ohm ? how about R_total...when you short the R₃ ...R_total also = 17ohm?

 

[vela]

You're not preserving the original circuit correctly with your transformations. For example, in your first picture, you have the R2/R5 combination connected directly to the battery, but in the original circuit, neither resistor is connected to the battery.

I split the original circuit into two pieces. Try replacing the portion on the left with its Norton equivalent. Also, since R2 and R5 are in series, you can replace them with one resistor. You should end up with a current source and three resistors in parallel.

 

[Loke]

mind if i see how you work it out the solution? i uploaded the pic...but if i shorted the R₃ ...then i don't know how to proceed.

 

[vela]

Ignore the righthand piece of the circuit completely for now and find the Norton equivalent of just the left piece.

 

[Loke]

yes... that's how i do in the beginning ...I_N=V₁₄/R₁₄=20/7=2.855A ...but why i still can't get 1.29A ...

 

[vela]

So what does the circuit look like after you've replaced the lefthand piece with its Norton equivalent?

 

[Loke]

ermmm..i don't know how to solve the lefthand piece with its norton equivalent...In=R25 ?

 

[vela]

You're not "solving" anything. Norton's theorem says you can replace a linear circuit with a current source in parallel with a resistor.

If you short the output of the lefthand piece, you will find I_N=20/7 A. Now, zero/short the voltage source and find the equivalent resistance of the lefthand piece. You will get R=7 ohms. Norton's theorem says you can replace the original voltage source and two resistors with the 20/7-A current source in parallel with the 7-ohm resistor.

If you draw the new circuit, you should have the current source, the 7-ohm resistor, R₃, and the combination of R₂ and R₅ all in parallel.

 

[Loke]

ermmm ...then what should i do next in order to get I₃=1.29A ? oh the second pic is the correct 1 ~?

 

[vela]

The second picture is close. Are you leaving R₃ out for some reason?

 

[Loke]

u meant R_N parallel with R₃ parallel with R₂₅ and parallel with current source ?

 

[vela]

Yes.

 

[Loke]

oh i got it already! find the total V and then find for I3

 

[vela]

No. You're trying to find the current in R₃, right? Try looking up "current divider."

 

[Loke]

Thanks ~~~~ i got it ^^

 

[Loke]

but when i use thevenin theorem much more easier !