Answer:

To balance the Wheatstone bridge network, the product of the resistances in one diagonal arm must be equal to the product of the resistances in the other diagonal arm. The formula for balancing a Wheatstone bridge is:

\[ R_1 \times R_4 = R_2 \times R_3 \]

In this case, let's assign the resistances as follows:

\( R_1 = 4 \, \Omega \)

\( R_2 = 4 \, \Omega \)

\( R_3 = 4 \, \Omega \)

\( R_4 = 10 \, \Omega \)

Now, substitute these values into the formula:

\[ 4 \times 10 = 4 \times R_2 \]

Solve for \( R_2 \):

\[ R_2 = \frac{4 \times 10}{4} \]

\[ R_2 = 10 \, \Omega \]

So, the resistance to be connected across the 12 Ω resistor to balance the network is \( \boxed{10 \, \Omega} \).