The Thevenin equivalent voltage, V_TH , in V (rounded off to 2 decimal places) of the shown below network, is _______.

The correct answer is 14.

Step-by-Step Explanation:

1. Understand the Circuit Diagram:
From the given circuit diagram, we can observe the following components:
• An independent DC voltage source of 4 V.
• A parallel resistor of 3 Ω connected directly across the 4 V voltage source.
• A series resistor of 2 Ω connected between the node above the voltage source and the middle node.
• An independent current source of 5 A connected to the middle node pointing upwards.
• A resistor of 3 Ω connected in series to the open-circuit output terminals across which the Thevenin equivalent voltage, $V_{TH}$, is to be measured.

2. Analyze the Open-Circuit Terminals:
Since the terminals for $V_{TH}$ are open-circuited, no current flows through the 3 Ω resistor connected to the positive open terminal.
Therefore, the current through the 3 Ω terminal resistor is zero amperes (0 A), and the voltage across it is 0 V.
Thus, the Thevenin voltage $V_{TH}$ is simply equal to the potential at the node between the 2 Ω resistor and the 5 A current source (taking the bottom reference wire as 0 V).

3. Calculate Node Voltage using Nodal Analysis:
Let the reference node (bottom wire) be at 0 V.
The node to the left of the 2 Ω resistor is directly connected across the 4 V voltage source, so its potential is fixed at 4 V.
Let $V_A$ be the potential at the node above the 5 A current source.

Applying Kirchhoff's Current Law (KCL) at node A (where current leaves the node):

$\frac{V_A-4}{2}-5=0$

4. Solve for $V_A$:

$\frac{V_A-4}{2}=5$

$V_A-4=10$

$V_A=14\text{V}$

Since the open-circuit output voltage is equal to $V_A$:

$V_{TH}=V_A=14\text{V}$

Thus, the Thevenin equivalent voltage $V_{TH}$ is 14 V.