Kirchoff's Laws

  • Thread starter Liquidxlax
  • Start date
  • #1
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Homework Statement



don't mind my cruddy drawing :P

WYPd5.jpg


R1 to R3 = 1000 Ω

R4 to R5 = 2000 Ω

i1 = 1mA= 1x10-3A

Find the current through each resistor.

Homework Equations



parallel Rp = (sum)(1/Ri)-1

series Rs = (sum)(Ri))

ΔV=iR


The Attempt at a Solution



i'm not sure of the best way to star this one because it is a little awkward to me. I thought the best way to do it would be to find Equivalent Resistance, but I have no idea how to do that with this geometry.

i know current at the start equals the end. each loop from start to finish ends in 0 potential


0 = ΔV -i2R1 - i5R2 - i6R3

0 = ΔV -i2R1 - i4R5

0= ΔV - i3R4 - i6R3

i1 = i2 + i3

thanks
 

Answers and Replies

  • #2
berkeman
Mentor
59,428
9,543

Homework Statement



don't mind my cruddy drawing :P

WYPd5.jpg


R1 to R3 = 1000 Ω

R4 to R5 = 2000 Ω

i1 = 1mA= 1x10-3A

Find the current through each resistor.

Homework Equations



parallel Rp = (sum)(1/Ri)-1

series Rs = (sum)(Ri))

ΔV=iR


The Attempt at a Solution



i'm not sure of the best way to star this one because it is a little awkward to me. I thought the best way to do it would be to find Equivalent Resistance, but I have no idea how to do that with this geometry.

i know current at the start equals the end. each loop from start to finish ends in 0 potential


0 = ΔV -i2R1 - i5R2 - i6R3

0 = ΔV -i2R1 - i4R5

0= ΔV - i3R4 - i6R3

i1 = i2 + i3

thanks

When in doubt, I just start writing KCL equations and then solve them for the unknowns.

So in this circuit, I'd draw ground at the right node, put an unknown voltage V at the left node, and write KCL equations for the middle 2 nodes. Solve away, and see if that gets you what you want...
 
  • #3
322
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if i'm thinking of the right nodes... how can you write an equation for the middle? you have to include the outer ones. i wrote out every equation i could possibly think of.

I'm to tired right now to fuss over the little things. just imagine the numbers are the currents and the nodes are a b c d

@ a 1 = 2 + 3

@ b 2 = 4 + 5

@ c 6 = 5 + 3

@ d 1 = 6 + 4

6+4 = 3+2

6+4= 6-5 +2

4 = 2-5

kind aggravated with this one
 
  • #4
berkeman
Mentor
59,428
9,543
if i'm thinking of the right nodes... how can you write an equation for the middle? you have to include the outer ones. i wrote out every equation i could possibly think of.

I'm to tired right now to fuss over the little things. just imagine the numbers are the currents and the nodes are a b c d

@ a 1 = 2 + 3

@ b 2 = 4 + 5

@ c 6 = 5 + 3

@ d 1 = 6 + 4

6+4 = 3+2

6+4= 6-5 +2

4 = 2-5

kind aggravated with this one

I'm not understanding your post. The KCL equations will be one per node, and will be in the form of a sum of currents. Each current will generally be expressed as a voltage difference between 2 nodes divided by the resistance between those two nodes.
 
  • #5
322
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I'm not understanding your post. The KCL equations will be one per node, and will be in the form of a sum of currents. Each current will generally be expressed as a voltage difference between 2 nodes divided by the resistance between those two nodes.

i guess i will redraw my picture and label the nodes. i know what you mean, by kcl. the first node would be

current 1 = current 2 + current 3
 
  • #6
berkeman
Mentor
59,428
9,543
i guess i will redraw my picture and label the nodes. i know what you mean, by kcl. the first node would be

current 1 = current 2 + current 3

I prefer to write my KCL equations as the sum of all currents leaving a node is zero. But that's up to personal preference.
 
  • #7
322
0
I prefer to write my KCL equations as the sum of all currents leaving a node is zero. But that's up to personal preference.

okay but at the nodes 3 and 4 with current i5 and i4, do those split up 2 ways or do they just continue to the end of the circuit?
 

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