KCL Question: Determine Currents

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The discussion revolves around solving a circuit problem using Kirchhoff's Current Law (KCL) and mesh analysis. The initial attempts to calculate the currents in the circuit branches were incorrect, leading to confusion about the application of KCL and the need for simultaneous equations. Participants emphasized the importance of combining series resistors and writing proper KCL equations for the circuit. After several iterations, a participant derived equations but still found discrepancies with the book's answers. Ultimately, the conversation highlighted the effectiveness of nodal analysis over KCL for this specific problem.
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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
Io1=\frac{16.00}{8.00}
=2.00A

Io2=\frac{16}{6}
=0.75A

Io3=\frac{16}{1}+\frac{16}{3}
=21.3A

P.S
 
Last edited by a moderator:
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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
Io1=\frac{16.00}{8.00}
=2.00A

Io2=\frac{16}{6}
=0.75A

Io3=\frac{16}{1}+\frac{16}{3}
=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:
Use mesh analysis and KVL
 
Zayer said:
Use mesh analysis and KVL

Is there a reason to pick KVL over KCL here? Just curious.
 
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.
 
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.
 
yes they are KCL equations
 
ok thanks i know how to do it now.
 

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