Homework Help: Proving Kirchoff's Laws algebraically

1. Feb 1, 2012

shyguy79

1. The problem statement, all variables and given/known data
From the attached file using Kirchoff's laws prove that V AB = V AD and that i1 = i/2

2. Relevant equations
V1 + V2 + V3 +... Vn = 0 (The sum of the voltage applied and dropped across the components is zero.
i1 + i2 = i3 (The sum of the branch currents is equal to the total current)

3. The attempt at a solution
I thought I was ok with algebraic manipulation but this has stumped me :-( any pointers would be appreciated as always!

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2. Feb 1, 2012

vela

Staff Emeritus
What did you get when you tried to apply Kirchoff's laws to the circuit?

3. Feb 1, 2012

shyguy79

Hi, as there are no figures given it looks like I'm expected to do prove everything using algebra...

I've calculated hypothetic values - say V = 12v, R total = 16Ω and i total = 0.75A and everything seems to add up but I'm afraid my algebra manipulation skills are somewhat limited.

For example I've tried that if: V ab = V ad and i1 = i - i2 and i2 = i -i1 (Kirchoffs 2nd Law) then i1 R0 = i2 R0 but don't know where to go now

4. Feb 1, 2012

vela

Staff Emeritus
You can't assume VAB=VAD because that's what you're trying to prove.

Applying KCL to node A gives you $i = i_1 + i_2$. I'm assuming all currents are flowing left to right. What do you get if you apply it to nodes B, C, and D?

5. Feb 1, 2012

shyguy79

B: i3 = i1
C: i = i3 + i4
D: i4 = i2

6. Feb 1, 2012

vela

Staff Emeritus
Great! So we can eliminate i3 and i4 because we know they're equal to i1 and i2 respectively, and the equation for node C tells us the same thing as the equation for node A.

Now apply KVL to the loop ABCDA. What do you get?

7. Feb 1, 2012

shyguy79

VT - i1R0 - i1R0 - i2R0 - i2R0 = 0
VT = VAB + VBC + VDC + VAD
VT = R0 (i1+i1+i2+i2)
VT = R0 (2i1 + 2i2)

But I don't understand how V AB can be described as V AD

or even

VT = VAB + VBC + VDC + VAD

Last edited: Feb 1, 2012
8. Feb 1, 2012

vela

Staff Emeritus
This isn't quite right. Across the four resistors, Ohm's Law tells us the potential differences are

VAB=i1R0
VBC=i1R0
VDC=i2R0

where the higher potential is at the left end of each resistor (where the current enters). So starting at A and going clockwise around the loop, we first encounter a voltage drop VAB and then another drop VBC, but going from C to D, the potential increases by VDC because we are going from right to left, and again from D to A by VAD. So KVL gives us

- VAB - VBC + VDC + VAD = 0

Note that VT doesn't enter into the picture because the battery is not part of the loop ABCDA. So plugging in the values above, you get

-i1R0-i1R0+i2R0-i2R0=0

9. Feb 1, 2012

shyguy79

Oh, I see! So the direction of flow has a direct effect on the sign of the variable! Cool, there are a few more questions in this assignment but I think I understand.

Thank you for all your help!