Solving Voltage Source Nodal Analysis: Find Vo

[STEMucator]

Homework Statement

Use nodal analysis to find $V_o$.

Homework Equations

$\sum I = 0$
$I = V/R$

The Attempt at a Solution

So the book started throwing voltage sources into nodal analysis and I'm a bit unsure about this problem. The answer is listed as $V_o = 3.89 V$.

All they have taught so far is to apply KCL to find the node voltages. Also, if a voltage source is connected between ground and a node, then the voltage at that node is equal to the voltage of the source.

This leads me to an important question about the problem at hand. I believe ground is located directly beneath the $12 V$ independent voltage. So couldn't I say directly above the voltage source and below the resistor is some $V_x = 12 V$?

Finding the voltage $V_2$ across the middle branch has proven to be troublesome.

 

[lazyaditya]

You do know how to apply KCL at nodes right. I am following these notations as you have not marked node voltages in your figure.

 

[lazyaditya]

Since we know the voltage below 6kohm resistor is 12 V and at the node we have considered V₂ in the middle branch (that is V₂ is the voltage across the whole middle branch).
Thus we can see that voltage drop across 6kohm resistor is (V₂ -12 ),now as we know the voltage drop we can calculate the current through 6kohm resistor by using ohm's law (Voltage drop/current).

 

[STEMucator]

You do know how to apply KCL at nodes right. I am following these notations as you have not marked node voltages in your figure.

Yes I know how to use KCL and I had those equations a while ago.

Since we know the voltage below 6kohm resistor is 12 V and at the node we have considered V₂ in the middle branch (that is V₂ is the voltage across the whole middle branch).
Thus we can see that voltage drop across 6kohm resistor is (V₂ -12 ),now as we know the voltage drop we can calculate the current through 6kohm resistor by using ohm's law (Voltage drop/current).

This post was helpful I was unsure about how to treat the $12 V$ source. I see now that we can use the difference between $V_2$ and the source to find the current across that branch.

The rest is easy.