# KCL Question

1. Feb 5, 2010

### Paymemoney

1. The problem statement, all variables and given/known data
Determine the current in each branch of the circuit in the below diagram.
http://img341.imageshack.us/img341/7901/kclquestion.jpg [Broken]

2. Relevant equations
V=IR
P=VI

3. 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: May 4, 2017
2. Feb 6, 2010

### Staff: Mentor

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: May 4, 2017
3. Feb 6, 2010

### Zayer

Use mesh analysis and KVL

4. Feb 6, 2010

### Staff: Mentor

Is there a reason to pick KVL over KCL here? Just curious.

5. Feb 6, 2010

### Paymemoney

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. Feb 6, 2010

### Staff: Mentor

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. Feb 6, 2010

### Paymemoney

yes they are KCL equations

8. Feb 7, 2010

9. Feb 7, 2010

### Paymemoney

ok thanks i know how to do it now.