Finding Currents Using Kirchhoff's Laws

[Angie K.]

Homework Statement

a. Determine the currents I1, I2, and I3 in the above figure. Assume the intermal resistance of each battery is r = 2.48Ω.

Homework Equations

Kirchhoff's Laws (Junction Rule and Loop Rule)

The Attempt at a Solution

Using Kirchhoff's Laws,
I ended up with three equations:
i1-i2-i3=0
-12i1-2.48i2=4.48
2.48i2-2.48i3=29

Using matrices to find i1, i2 and i3 I come up with answers but they are wrong. I am wondering if there is a mistake that I made and am not seeing it?

 

[gneill]

Your KCL equation looks okay, but your KVL equations don't look right. Can you give more detail about how you arrived at them? Perhaps write them out showing all contributing terms as you do your "KVL walk" around the loop.

 

[Doc Al]

i1-i2-i3=0

This looks good.

-12i1-2.48i2=4.48
2.48i2-2.48i3=29

How did you get these two equations?

 

[Angie K.]

The Loop Rule Equations:

(Starting from the top, to the left of the 12V battery and going counterclockwise) -2.48-8+12-2.48i2-10-12i+12=0

(Starting from the junction of the right) 10+2.48i2-12-15+6-2.48i3-18=0

 

[Doc Al]

(Starting from the top, to the left of the 12V battery and going counterclockwise) -2.48-8+12-2.48i2-10-12i+12=0

For one thing, you forgot to multiply the internal resistance by the current to get the voltage drop.

 

[gneill]

The Loop Rule Equations:

(Starting from the top, to the left of the 12V battery and going counterclockwise) -2.48-8+12-2.48i2-10-12i+12=0

(Starting from the junction of the right) 10+2.48i2-12-15+6-2.48i3-18=0

I'm seeing a mix of voltage and resistance terms, some resistances multiplied by currents and some not. So as written those equations cannot be correct. You are summing potential changes around the loop, so any resistor value must be multiplied by a current in order to realize its potential change (Ohm's Law). Perhaps you could group the resistance terms for given currents in parentheses?

 

[Angie K.]

The Loop Rule Equations:

(Starting from the top, to the left of the 12V battery and going counterclockwise) -2.48-8+12-(2.48i2)-10-(12i1)+12=0

(Starting from the junction of the right) 10+(2.48i2)-12-15+6-(2.48i3)-18=0

 

[gneill]

(Starting from the top, to the left of the 12V battery and going counterclockwise) -2.48-8+12-(2.48i2)-10-(12i1)+12=0

Aren't 2.48 and 8 and 10 all resistances? What currents multiply them? I see only two voltage sources in the loop (both 12 V).

(Starting from the junction of the right) 10+(2.48i2)-12-15+6-(2.48i3)-18=0

Again, I don't see currents multiplying each resistance.

 

[Angie K.]

The Loop Rule Equations:

(Starting from the top, to the left of the 12V battery and going counterclockwise) -2.48i1-8i1+12-2.48i2-10i2-12i+12=0

(Starting from the junction of the right) 10i2+2.48i2-12-15i3+6-2.48i3-18i3=0

 

[gneill]

That looks better! What do you find for your currents?

 

[Angie K.]

Got it, thanks! I just needed to be consistent with labeling which current goes with which resistor.