In the circuit shown below, the magnitude of the voltage V₁ in volts, across the 8kΩ resistor is _______. (round off to nearest integer)

The correct answer is 100.

Circuit Analysis:
Based on the circuit diagram:

The circuit consists of:
- An independent DC voltage source of 75 V.
- A 2 kΩ resistor in the upper-left branch, through which a current I flows to the right.
- A voltage-controlled dependent voltage source of value 0.5 V₁.
- A central vertical path which connects the node after the dependent voltage source directly to the reference (ground) node at the bottom.
- A current-controlled dependent current source of value I, pointing to the right.
- An 8 kΩ resistor in the rightmost branch, across which the voltage is V₁ (with the positive terminal at the top).

Step 1: Analyzing the Left Loop using Kirchhoff's Voltage Law (KVL)
Because the node between the dependent voltage source and the dependent current source is connected to the common ground wire via the central vertical short-circuit, the voltage at this node is 0 V.
Applying KVL to this closed loop on the left:

$75-I·\left(2\text{kΩ}\right)-0.5V_1=0$

Rearranging this to solve for the current I:

$I=\frac{75-0.5V_1}{2000}$

Step 2: Analyzing the Right Loop
The dependent current source is connected in series with the 8 kΩ resistor, meaning it dictates the current flowing through it. A current of I flows downwards through the 8 kΩ resistor.
Using Ohm's Law to relate the voltage V₁ to the current I:

$V_1=I·\left(8\text{kΩ}\right)=8000I$

Step 3: Solving for V₁
Substitute the expression for I from Step 1 into the equation in Step 2:

$V_1=8000·\left(\frac{75-0.5V_1}{2000}\right)$

Simplifying the fraction:

$V_1=4·(75-0.5V_1)$

Expanding the brackets:

$V_1=300-2V_1$

Adding 2V₁ to both sides:

$3V_1=300$

$V_1=100\text{V}$

Thus, the magnitude of the voltage V₁ across the 8 kΩ resistor is 100 V.