Solve Kirchoff's Laws Homework: R1-R5, i1=1mA

  • Thread starter Liquidxlax
  • Start date
  • Tags
    Laws
In summary: 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?
  • #1
322
0

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 potential0 = ΔV -i2R1 - i5R2 - i6R3

0 = ΔV -i2R1 - i4R5

0= ΔV - i3R4 - i6R3

i1 = i2 + i3

thanks
 
Physics news on Phys.org
  • #2
Liquidxlax said:

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
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
Liquidxlax said:
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
berkeman said:
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
Liquidxlax said:
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
berkeman said:
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?
 

1. What are Kirchoff's Laws?

Kirchoff's Laws are a set of fundamental principles in electrical circuit analysis. They include Kirchoff's Current Law (KCL), which states that the sum of currents entering a node must equal the sum of currents leaving the node, and Kirchoff's Voltage Law (KVL), which states that the sum of voltage drops around a closed loop must equal the sum of voltage rises.

2. How do I apply Kirchoff's Laws to solve a circuit?

To apply Kirchoff's Laws, you must first identify all the current and voltage sources in the circuit. Then, you can use KCL and KVL to write equations for each node and loop in the circuit. Finally, you can solve these equations simultaneously to find the values of the unknown currents and voltages.

3. What is the formula for KCL?

The formula for KCL is: ∑iin = ∑iout, where ∑iin is the sum of currents entering a node and ∑iout is the sum of currents leaving the node.

4. How do I determine the direction of current in KCL equations?

In KCL equations, the direction of current is typically assumed to be flowing into the node. If it is actually flowing out of the node, the value for that current will be negative in the equation.

5. Can Kirchoff's Laws be used for any type of circuit?

Yes, Kirchoff's Laws can be applied to any type of circuit, including series, parallel, and complex circuits. However, they are most commonly used for DC circuits and may need to be modified for AC circuits.

Suggested for: Solve Kirchoff's Laws Homework: R1-R5, i1=1mA

Back
Top