KCL Question: Determine Currents

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In summary, the conversation discusses a problem in which the current in each branch of a circuit needs to be determined. The equations V=IR and P=VI are mentioned, and the solution involves using KCL and mesh or nodal analysis to solve for the currents. The correct solution is found using nodal analysis.
  • #1
Paymemoney
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Homework Statement


Determine the current in each branch of the circuit in the below diagram.
http://img341.imageshack.us/img341/7901/kclquestion.jpg

Homework Equations


V=IR
P=VI

The Attempt at a Solution


This is what i have done but it is incorrect.
Vtotal=16V
V=IR
[tex]Io1=\frac{16.00}{8.00}[/tex]
=2.00A

[tex]Io2=\frac{16}{6}[/tex]
=0.75A

[tex]Io3=\frac{16}{1}+\frac{16}{3}[/tex]
=21.3A

P.S
 
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  • #2
Paymemoney said:

Homework Statement


Determine the current in each branch of the circuit in the below diagram.
http://img341.imageshack.us/img341/7901/kclquestion.jpg

Homework Equations


V=IR
P=VI

The Attempt at a Solution


This is what i have done but it is incorrect.
Vtotal=16V
V=IR
[tex]Io1=\frac{16.00}{8.00}[/tex]
=2.00A

[tex]Io2=\frac{16}{6}[/tex]
=0.75A

[tex]Io3=\frac{16}{1}+\frac{16}{3}[/tex]
=21.3A

P.S

You are not writing the simultaneous KCL equations as far as I can tell. Combine series resistors where you can, and write the 1 KCL equation for this circuit (why just 1?). Then solve it.
 
Last edited by a moderator:
  • #3
Use mesh analysis and KVL
 
  • #4
Zayer said:
Use mesh analysis and KVL

Is there a reason to pick KVL over KCL here? Just curious.
 
  • #5
After redoing the question this is what i have done:

Three equations:

Equation 1: Io3=Io1+1o2

Equation 2: 8-(4)Io2-(6)Io3=0

Equation 3: 4-(8)Io1-(6)Io2=0

Substitute Equation 1 into Equation 2
Equation 4
8-(4)Io2-(6)(Io1+Io2)=0
8=(10)Io2+(6)Io1

Equation 4 - Equation 3

8=(10)Io2+(6)Io1 - 4=(8)Io1+(6)Io2

Io2=0.90Amps

Substitute Io2 into Equation 3

Io1=\frac{4-(6*0.90)}{8}

=0.175Amps
Subsitute Io1 and Io2 into Equation 1

Io3=Io1+Io2
=0.90+0.175
=1.075Amps

I checked with the book's answers and my solutions are incorrect. Can someone tell me where i have gone wrong.
 
  • #6
I do not understand your equations. Are they meant to be KCL equations? The KCL only involves the voltage at that top node, and the 3 currents that flow out of that node.
 
  • #7
yes they are KCL equations
 
  • #9
ok thanks i know how to do it now.
 

FAQ: KCL Question: Determine Currents

What is KCL (Kirchhoff's Current Law)?

KCL is a fundamental law in circuit analysis that states that the sum of all currents entering a node (or junction) in a circuit is equal to the sum of all currents leaving that node. In other words, the total current flowing into a node must equal the total current flowing out of that node.

Why is KCL important in circuit analysis?

KCL is important because it allows us to determine the unknown currents in a circuit by using a simple conservation of charge principle. It is a powerful tool for analyzing complex circuits and is essential for solving many real-world engineering problems.

What is the difference between KCL and KVL (Kirchhoff's Voltage Law)?

KCL deals with the currents in a circuit, while KVL deals with the voltages. KCL is based on the principle of conservation of charge, while KVL is based on the principle of conservation of energy. In simple terms, KCL is about the flow of charge, while KVL is about the difference in potential between two points in a circuit.

How do you apply KCL to a circuit?

To apply KCL, you must first identify all the nodes (junctions) in the circuit. Then, write an equation for each node by summing up all the currents entering and leaving that node. Finally, solve the resulting system of equations to determine the unknown currents in the circuit.

Are there any limitations to KCL?

Yes, there are a few limitations to KCL. It assumes that the circuit is in a steady state (i.e. all voltages and currents are constant) and that there are no capacitors or inductors present. Additionally, it only applies to circuits with a single closed loop (or mesh).

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