Question

There is an infinite wire grid with cells in the form of equilateral triangles. The resistance of each wire between neighboring joint connections is $R_0$. The net resistance of the whole grid between the points $A$ and $B$ as shown in Figure is

• A. $R_0$
• B. $\frac{R_0}{2}$
• C. $\frac{R_0}{3}$
• D. $\frac{R_0}{4}$

Answer 1

The correct option is C $\frac{R_0}{3}$

We will solve this problem by the superposition principle. Let current $I$ enter at A, then current $I/6$ will flow through AB. Now, in the second case, let the current be extracted from B, again current $I/6$ will flow through AB. Now, if simultaneously current enters at A and leaves at B, then the current $I/6+I/6=I/3$ should flow through AB and current $I-I/3=2I/3$ will flow through the remaining part. Let the resistance of the remaining part be $R_1$. Now the circuit is shown as in figure.
As $R_0$ and $R_1$ are in parallel so
potential across $R_0=$ potential across $R_1$.
or $\left(I/3\right)R_0=\left(2I/3\right)R_1\Rightarrow R_1=R_0/2$
Thus the equivalent resistance $R_{eq}=\frac{R_0{R}_1}{R_0+R_1}=\frac{R_0\left(R_0/2\right)}{R_0+R_0/2}=R_0/3$