The output voltage V (in Volt) for the network given in the Figure is . (rounded off to two decimal places)

The correct answer is 1.00.

Step-by-step Explanation:

Let us analyze the given circuit diagram by labeling the nodes and applying Kirchhoff's Current Law (KCL):
1. Let the bottom wire be the reference node (Ground, $V_{\text{ref}}=0\text{V}$).
2. The top wire is at the potential $V_o$ with respect to the reference node, which represents the output voltage we need to find.
3. Let the node between the two left resistors be node $A$ at node voltage $V_A$.
4. Let the node between the two right resistors be node $B$ at node voltage $V_B$.
5. A $6\text{V}$ independent voltage source is connected between node $A$ and node $B$ with the positive terminal at node $A$, giving:

$V_A-V_B=6\text{V}$

1. Applying KCL at the bottom node (Ground, 0 V):
The current source of $1\text{mA}$ is connected between the bottom wire and the top wire, drawing current out of the bottom node. The two bottom $1\text{k}\Omega$ resistors are connected to this reference node. Applying KCL at the bottom node:

$\frac{V_A}{1\text{k}\Omega }+\frac{V_B}{1\text{k}\Omega }=1\text{mA}$

Multiplying the entire equation by $1\text{k}\Omega$:

$V_A+V_B=1\text{V}$

2. Applying KCL at the top node (Vo):
The current source injects $1\text{mA}$ into the top node. The current leaves the top node through the two top $1\text{k}\Omega$ resistors connected to nodes $A$ and $B$:

$1\text{mA}=\frac{V_o-V_A}{1\text{k}\Omega }+\frac{V_o-V_B}{1\text{k}\Omega }$

Multiplying by $1\text{k}\Omega$:

$1=V_o-V_A+V_o-V_B$

Rearranging the terms:

$2V_o-(V_A+V_B)=1$

$2V_o=V_A+V_B+1$

3. Solving for Vo:
Substitute the value of $V_A+V_B=1\text{V}$ from KCL at the bottom node into the top node equation:

$2V_o=1+1$

$2V_o=2$

$V_o=1.00\text{V}$

Thus, the output voltage $V_o$ is exactly $1.00\text{V}$.