In the given circuit, for voltage V_y to be zero, the value of β should be ________.(Round off to 2 decimal places)

The correct answer is -3.25.

Here is the step-by-step analysis of the circuit to determine the value of $\beta$ for which the voltage $V_y$ becomes zero:

Step 1: Apply Kirchhoff's Current Law (KCL) at node x
We assume all currents are leaving node x:

$\frac{V_x-6}{1}+\frac{V_x}{4}+\frac{V_x-V_y}{2}=0$
To clear the fractions, multiply the entire equation by 4:

$4(V_x-6)+V_x+2(V_x-V_y)=0$
$4V_x-24+V_x+2V_x-2V_y=0$
$7V_x-2V_y=24$ (Equation 1)

Step 2: Find the value of $V_x$ when $V_y=0\text{V}$
Substituting $V_y=0$ into Equation 1:

$7V_x-2\left(0\right)=24$
$V_x=\frac{24}{7}\text{V}$

Step 3: Apply Kirchhoff's Current Law (KCL) at node y
We assume all currents are leaving node y. Note that the 2 A current source is entering node y, so its leaving current is -2 A:

$\frac{V_y-V_x}{2}-2+\frac{V_y-\beta V_x}{3}=0$
Multiply the entire equation by 6 to clear the fractions:

$3(V_y-V_x)-12+2(V_y-\beta V_x)=0$
$3V_y-3V_x-12+2V_y-2\beta V_x=0$
Group the terms:

$5V_y-(3+2\beta )V_x=12$ (Equation 2)

Step 4: Solve for $\beta$
Substitute $V_y=0$ and $V_x=\frac{24}{7}$ into Equation 2:

$5\left(0\right)-(3+2\beta )\left(\frac{24}{7}\right)=12$
Divide both sides by 12:

$-(3+2\beta )\left(\frac{2}{7}\right)=1$
Multiply both sides by $-\frac{7}{2}$:

$3+2\beta =-3.5$
Subtract 3 from both sides:

$2\beta =-6.5$
Divide by 2:

$\beta =-3.25$