Kirchoffs Current Law to find voltage?

In summary: The negative current means i is flowing the opposite way to what I have marked (towards top of...It means i is flowing towards the bottom of the circuit.
  • #1
TheNarrator
9
0

Homework Statement


Screen_Shot_2015_05_14_at_5_09_25_pm.png

Am having trouble solving the initial part of this problem, I think I am being confused by the dual voltage sources.

2. Homework Equations

I=V/R ΣI at a node =0

The Attempt at a Solution


V3/R3 - (V1/R1+V2/R2) = 0
V3=12-VA
V1=4-VA
V2=VA-4

However this must be incorrect somewhere as It is impossible to isolate Va from the above equations. Am normally very strong mathematically but find basic electronics very confusing for some reason!
 
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  • #2
TheNarrator said:

Homework Statement


Screen_Shot_2015_05_14_at_5_09_25_pm.png

Am having trouble solving the initial part of this problem, I think I am being confused by the dual voltage sources.

2. Homework Equations

I=V/R ΣI at a node =0

The Attempt at a Solution


V3/R3 - (V1/R1+V2/R2) = 0
V3=12-VA
V1=4-VA
V2=VA-4

However this must be incorrect somewhere as It is impossible to isolate Va from the above equations. Am normally very strong mathematically but find basic electronics very confusing for some reason!
Apply KCL for the node A: that means you express the currents through the resistors R1, R2, R3 in terms of the voltages VA, V1, V2, and then apply the nodal law. Remember, the current flowing through a resistor is proportional to the voltage difference between its terminals.
 
  • #3
ehild said:
Apply KCL for the node A: that means you express the currents through the resistors R1, R2, R3 in terms of the voltages VA, V1, V2, and then apply the nodal law. Remember, the current flowing through a resistor is proportional to the voltage difference between its terminals.
I understand that part, however I thought that the equations I had put in my attempted solution were in terms of voltage differences and the three voltages? I thought that the way the circuit was laid out, all three voltage differences would be dependant on Va?
 
  • #4
Yes. And KCL provides an equation for VA. Do as it is said in the problem.
So what is I1, the current through R1?
What is I2, the current through R2?
What is I, the current through R3?
 
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  • #5
TheNarrator said:
I understand that part, however I thought that the equations I had put in my attempted solution were in terms of voltage differences and the three voltages?

You didn't mark up the circuit so it's very hard to work out what you are doing. It doesn't look right. You have terms like V2/R2 which doesn't make sense. V2 is the 12V source on the right hand branch. R2 is a resistor in the middle branch.

Do what others have suggested. Start by marking the circuit with the currents I1, I2, I3 and any voltages you need.

Apply KCL but don't skip the very first step which is to sum the currents. There should be an equation of the form

Ia + Ib + Ic... = 0

Don't go straight to equations with V/R in them.
 
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  • #6
Just to add...

It's extremely easy to make trivial sign errors when applying KVC/KCL. It's essential to mark up the diagram because you need to refer to it at the end. For example if a voltage or current turns out to be negative you need to know which direction you defined as positive.
 
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  • #7
Thanks for your replies everyone, just at work atm but will mark up the diagram and post my current calculations tonight
 
  • #8
Marked_circuit.png

This shows what I have attempted, I think where I am going wrong is the way I am doing the voltage differences but I am really struggling to understand specifically where I am going wrong
 
  • #9
The currents are right , write up the Nodal Law: them current entering the node is equal to the current leaving it.
 
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  • #10
Okay cool, so if i1+i2+i=0

Then (-Va/8)+((Va-4)/6)+((Va-12)/4)=0

But I can't work out how to solve for Va from here, is there another equation I am meant to be using to solve with substitution?
 
  • #11
TheNarrator said:
Okay cool, so if i1+i2+i=0

Then (-Va/8)+((Va-4)/6)+((Va-12)/4)=0

But I can't work out how to solve for Va from here, is there another equation I am meant to be using to solve with substitution?
Va is the only unknown. Just collect the terms with Va.
 
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  • #12
Oh of course! brain fade..

I calculated Va as 12.57V so therefore i has a value of (12.57-12)/4 = .143A

Thanks for your help!

I will attempt to solve the thevenins problem by myself but will post later if I can't get there
 
  • #13
TheNarrator said:
Oh of course! brain fade..

I calculated Va as 12.57V so therefore i has a value of (12.57-12)/4 = .143A
Check it once more. I1 goes into the node, but the other two leave it.
 
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  • #14
ehild said:
Check it once more. I1 goes into the node, but the other two leave it.
Okay so now I have (-Va/8)-(Va/6)-(Va/4)=(-4/6)-(12/4)

Which gives Va as 6.77V and a corresponding i of -1.31A

The negative current means i is flowing the opposite way to what I have marked (towards top of diagram)
 
  • #15
TheNarrator said:
Okay cool, so if i1+i2+i=0

You didn't define if +ve is into or out of the node.

Is that equation consistent with your diagram (which defines some as +ve going into the Va node and some as leaving)?
 
  • #16
If you define +ve as current going into the node then I make it...

I1 + (-I2) + (-I) = 0
or
I1 - I2 - I =0
 
  • #17
TheNarrator said:
Okay so now I have (-Va/8)-(Va/6)-(Va/4)=(-4/6)-(12/4)

Which gives Va as 6.77V and a corresponding i of -1.31A

The negative current means i is flowing the opposite way to what I have marked (towards top of diagram)
Correct! Well done!
 

FAQ: Kirchoffs Current Law to find voltage?

What is Kirchoff's Current Law?

Kirchoff's Current Law, also known as Kirchoff's First Law, states that the sum of all currents entering and exiting a node (or junction) in a circuit must equal zero.

How is Kirchoff's Current Law applied to find voltage?

Kirchoff's Current Law is applied by using it to set up equations that relate the currents in the circuit to the known voltage sources. By solving these equations, the voltage at each node can be determined.

Can Kirchoff's Current Law be used in any circuit?

Yes, Kirchoff's Current Law can be applied to any circuit as long as it is a closed loop or has defined nodes and branches.

What are the limitations of Kirchoff's Current Law?

Kirchoff's Current Law assumes that the circuit is in a steady state and that the resistances in the circuit are constant. It also does not take into account any capacitance or inductance in the circuit.

How does Kirchoff's Current Law relate to conservation of charge?

Kirchoff's Current Law is based on the principle of conservation of charge, which states that charge cannot be created or destroyed. This law ensures that the total charge entering a node must equal the total charge exiting the node, thus maintaining the conservation of charge in a circuit.

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