
In the given arrangement, we observe a Wheatstone Bridge.
For the bridge to be balanced:

\[ \frac{V_{AB}}{V_{AD}} = \frac{V_{BC}}{V_{CD}}. \]
Substituting values from the circuit:
\[ \frac{12}{6 + x} = \frac{0.5}{0.5}. \]
Simplify:
\[ 12(0.5) = (6 + x)(0.5). \]
\[ 6 = 3 + 0.5x \implies x = 6 \, \Omega. \]
Final Answer: \( x = 6 \, \Omega \).
A square loop of sides \( a = 1 \, {m} \) is held normally in front of a point charge \( q = 1 \, {C} \). The flux of the electric field through the shaded region is \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is: