# Kirchhoff's law

1. Sep 17, 2010

### 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 IN=2.855A=I3 but when i use kirchhoff's law i get I3=0.645A.
-i dont know where is the problems...the current for norton theorem and kirchhoff's law should be the same right?

#### Attached Files:

• ###### kirchhoff's law.jpg
File size:
11.2 KB
Views:
96
Last edited: Sep 17, 2010
2. Sep 18, 2010

### vela

Staff Emeritus
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.

3. Sep 19, 2010

### Loke

attempt solution:
-norton theorem
IN=VS/(R1+R4)=2.86A

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

4. Sep 19, 2010

### vela

Staff Emeritus
What exactly is I3 supposed to be? And why would you expect it to be equal to IN?

5. Sep 19, 2010

### 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.

Last edited: Sep 19, 2010
6. Sep 19, 2010

### Loke

by the way, can you show me your solution? actually this is not my homework... i just wanna 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...

Last edited: Sep 19, 2010
7. Sep 19, 2010

### vela

Staff Emeritus
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.

8. Sep 19, 2010

### Loke

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

#### Attached Files:

File size:
7.7 KB
Views:
60
File size:
6.3 KB
Views:
52
• ###### Rtotal.jpg
File size:
8 KB
Views:
55
Last edited: Sep 19, 2010
9. Sep 19, 2010

### vela

Staff Emeritus
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.

#### Attached Files:

• ###### kirchhoff's law.jpg
File size:
5.9 KB
Views:
49
10. Sep 19, 2010

### Loke

mind if i see how you work it out the solution? i uploaded the pic....but if i shorted the R3 .....then i dont know how to proceed.

#### Attached Files:

• ###### 1.jpg
File size:
8.5 KB
Views:
70
11. Sep 20, 2010

### vela

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

12. Sep 20, 2010

### Loke

yes.... that's how i do in the beginning ......IN=V14/R14=20/7=2.855A ...but why i still cant get 1.29A ....

13. Sep 20, 2010

### vela

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

14. Sep 20, 2010

### Loke

ermmm..i dont know how to solve the lefthand piece with its norton equivalent....In=R25 ?

Last edited: Sep 20, 2010
15. Sep 20, 2010

### vela

Staff Emeritus
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 IN=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.

16. Sep 20, 2010

### Loke

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

#### Attached Files:

File size:
6.8 KB
Views:
54
• ###### 3.png
File size:
7.4 KB
Views:
59
17. Sep 20, 2010

### vela

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

18. Sep 20, 2010

### Loke

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

19. Sep 20, 2010

### vela

Staff Emeritus
Yes.

20. Sep 20, 2010

### Loke

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

Last edited: Sep 20, 2010