Kirchhoff's Current Law: Solving for Open Circuit Voltage

[princejan7]

Homework Statement

http://postimg.org/image/61e06uo7x/

The question is to find the open circuit voltage

Homework Equations

The Attempt at a Solution

Just wondering why the current flowing through the voltage source was left out for node 1
and the current through the 16 ohm resistor was left out for node 2?

 

[CWatters]

Just wondering why the current flowing through the voltage source was left out for node 1

It wasn't left out. That's the (24-V1)/10 term.

and the current through the 16 ohm resistor was left out for node 2?

It wasn't left out either. If the output is open circuit then the current flowing through the 16R is 3A.

Aside: Personally I don't like the way they wrote the equations. KCL basically states that the sum of the currents equals zero so I prefer to write the equations in the form I1+I2+I3=0 rather than I1+I2 = -I3

 

[princejan7]

It wasn't left out. That's the (24-V1)/10 term.



If the output is open circuit then the current flowing through the 16R is 3A.

can you explain why it is 3A
Doesn't the current source send current to the left so that the 16 resistor gets something less than 3A?

 

[SammyS]

can you explain why it is 3A
Doesn't the current source send current to the left so that the 16 resistor gets something less than 3A?

Doesn't all the current coming to the current source need to flow through the 16 Ω resistor?

 

[CWatters]

can you explain why it is 3A
Doesn't the current source send current to the left so that the 16 resistor gets something less than 3A?

No.

Apply KCL at the Vth+ node. 3A leaves the node via the current source so 3A must also enter the node from somewhere or the sum won't be zero. The only place it can come from is through the 16R.