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.

 

[rezeew]

Yes, you are correct in thinking that the voltage at the node directly above the voltage source is 12 V. This is because the voltage source is connected between that node and ground, so the voltage at that node must be equal to the voltage of the source.

To find the voltage at node 2, you can use Kirchhoff's Voltage Law (KVL) to set up an equation for the voltage around the loop containing the voltage source, resistor, and node 2. This equation would be: $V_2 + 12 V - 10 V = 0$, since the voltage across the resistor is equal to the voltage drop across the voltage source (since they are connected in series). Solving for $V_2$, we get $V_2 = -2 V$.

Now, we can use KCL to set up an equation for the current at node 2: $I_2 = (12 V - V_2)/10 Ω = (12 V - (-2 V))/10 Ω = 1.4 A$.

Finally, we can use Ohm's Law to find the voltage across the resistor: $V_R = I_2 * R = 1.4 A * 5 Ω = 7 V$.

Since the voltage at node 2 is -2 V and the voltage across the resistor is 7 V, we can use KVL to find the voltage at node 1: $V_1 + 7 V - 10 V = 0$. Solving for $V_1$, we get $V_1 = 3 V$.

Therefore, the voltage at node 1 (which is also $V_o$) is 3 V, which is close to the answer of 3.89 V listed in the book. This difference could be due to rounding or small errors in calculations.

In summary, nodal analysis is a powerful tool for solving circuits with multiple voltage sources. By using KVL and KCL, we can set up equations to solve for the unknown node voltages and currents. Remember to carefully consider the direction of current flow and the polarity of voltage sources when setting up these equations.