# Homework Help: Kirchhoff's circuit law

1. Aug 3, 2010

### Thrax

Hi,
I’m having some trouble with Kirchhoff's circuit law problems regarding two batteries in the circuit. I've attached a crude picture of a problem i picked to understand whats going on.
So for the bottom loop I have
5V - 5v = i1 - i2
0= i1 – i2
The top loop I’m coming up with
5 = i2 -2(i1+i2)
5= i2 -2i1 -2i2
-1(5 = -2i1 -i2)
So I’m left with
5 =2i1 +i2
0= i1 – i2
5=-2i1
0=- i1
5=-3i1
5/-3 = i1
But, I think this is wrong.

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2. Aug 4, 2010

### ehild

What convention do you use for the terminals of a battery? Is the short line positive or negative? If the short line is positive than the eq. for the top circuit is 5 = i2 + 2(i1+i2). (I1+I2) flows according to the "walking round" direction in the loop.

ehild

3. Aug 4, 2010

### Thrax

Well, I use whatever gets me there...But, the short line is negative.

4. Aug 4, 2010

### Thrax

I'm not sure what to do when i hit a battery. I know that you assign a positive value to the battery if the battery output follows the tracing direction. But, for some reason it dosnt always work out. Like you see in this problem....

5. Aug 4, 2010

### ehild

Follow the change of potential along the tracing direction in a loop. Choose a point like A with zero potential in the figure and trace the potential along the loop. It should be zero when returning to A.

For the bottom loop: According to the tracing direction, you go down then to the left. Passing the battery, the potential changes to -5 V. Along the first resistor, the potential falls by I1*R1, so you have -5-I1. Then you go to the right , along the second resistor, but the current flows against the tracing direction. The right end of the resistor is positive with respect to the left one: the potential becomes -5-I1+I2. Then the second battery follows, the potential increases in the tracing direction and becomes 0 at A:

-5-I1+I2+5=0.

For the top loop, you start at A again and follow the tracing direction. The potential drops by 5 V through the battery, than by I2*1 through the 1 ohm resistor, then by 2(I1+I2) through the 2 ohm resistor on the top.

-5-I2-2(I1+I2)=0

ehild

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6. Aug 4, 2010

### Thrax

Yes, I believe i do. Let me take a stab at it and see what trouble i can get into...

7. Aug 4, 2010

### ehild

What do you mean with that stab?

Solve the equations. What do you get?

ehild

8. Aug 4, 2010

### Thrax

I've got -1A for I1, 1A for I2 and 2A for I3

9. Aug 4, 2010

### Thrax

Sorry, typed that in wrong...I've got -1A for I2, 1A for I1 and 0? for I3.

10. Aug 4, 2010

### ehild

See the first equation: It is I1-I2=0. So I1=I2, is not it??

ehild

11. Aug 4, 2010

### Thrax

I3 is 2A, but i dont see the I1-I2 in this situation. Arnt I1 and I2 both converving onto
I3? Or Im I looking at it wrong?

12. Aug 4, 2010

### Thrax

Oh yeah...i see it now. Thanks!!!!!!

13. Aug 4, 2010

### Thrax

One more thing...This might be a stupid question. But once you've found I1 and I2. If one of those is a negative number...it indicates that the current is flowing in the other direction. So, if I3 = I1 + I2 and let’s say that I1=-1, and I2 = 1would that change the equation to
I3= I1-I2 giving you the 2A?

14. Aug 4, 2010

### ehild

The sign of the current can be anything it does not change the equation I3=I1+I2. For example, if f I1=2 A and I2 = -3 A, then I3 =(2)+(-3)= -1 A.

I1=I2 from the first equation. And I3=I1+I2 because of the nodal rule. Is it all right?

Plug in I1 for I2 in the second equation.

-5-I2-2(I1+I2)=0 --->-5-I1-2(I1+I1)=0 --->-5-5I1=0 --->

I1=-1A, I2=-1A. I3=I1+I2=(-1)+(-1) = -2 A. .

So all currents flow in the opposite direction as you assumed.

ehild

15. Aug 4, 2010

### Delta2

You make the equations assuming that the currents flow to some directions. After you solve the equations a minus sign just tells you that the current flows to opposite direction than the one you assumed. Now that you know the real current flows if you make the equations again they will be different BUT if you solve the new equations you will not find minus signs.