Twelve resistors each of resistance 16 ,Omega | vedclass

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(D) The circuit consists of three concentric squares,each having four resistors. Let the resistance of each resistor be $R = 16 \,\Omega$.$1$. The inn

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(D) The circuit consists of three concentric squares,each having four resistors. Let the resistance of each resistor be $R = 16 \,\Omega$.$1$. The innermost square has four resistors in series,but they are connected to the next square. By analyzing the symmetry,we can simplify the circuit by grouping resistors in parallel.$2$. Each side of the three squares consists of resistors. When we look at the circuit from $A$ to $B$,we can simplify the network into an equivalent circuit where each branch has an effective resistance of $R/3$.$3$. The simplified circuit consists of a Wheatstone bridge-like structure or a parallel-series combination where the equivalent resistance $R_{eq}$ between $A$ and $B$ is given by the parallel combination of two branches: one branch with resistance $R/3$ and another branch with resistance $R/3 + R/3 + R/3 = R$.$4$. Calculating the equivalent resistance: $R_{eq} = \frac{(R/3) 	imes R}{(R/3) + R} = \frac{R^2/3}{4R/3} = \frac{R}{4}$.$5$. Given $R = 16 \,\Omega$,the net resistance is $R_{eq} = \frac{16}{4} = 4 \,\Omega$.

Twelve resistors each of resistance $16 \,\Omega$ are connected in the circuit as shown. The net resistance between $A$ and $B$ (any two conjugate points of the outer square) is ............... $\Omega$.

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