Question

The resistance of all the wires between any two adjacent dots is R. The equivalent resistance between A and B as shown in the figure is

• A. $\frac{7}{3}R$
• B. $\frac{7}{6}R$
• C. $\frac{14}{8}R$
• D. None of these

Answer 1

The correct option is B $\frac{7}{6}R$

Since we need to evaluate the resistance between A and B, therefore connect a power source between A and B as shown below.

Since point D and E are connected to A by a wire of equal resistance, therefore both points D and E will be at equal potential. Hence we can neglect the resistance between D and E since no current will flow through it.
The circuit can be redrawn as shown.
For the branches CDG, EFH, GBI and BHJ, two equal resistances($R$) are in series and their equivalent($2R$) is in parallel with the third resistor($R$).
Hence the equivalent resistance for each branch will be
$R_{eq1}=\frac{2R\times R}{2R+R}=\frac{2R}{3}$
Hence the circuit reduces to the following circuit.
Now the circuit is symmeteric about AB. For one side of AB say left, three resistances $R$, $2R/3$ and $2R/3$ are in series. Their equivalent resistance will be,
$R_{eq2}=R+\frac{2R}{3}+\frac{2R}{3}=\frac{7R}{3}$
Hence the circuit reduce to the one as shown below.
Since two equal resistance are connected across AB in parallel, there equivalent resistance will be,
$R_{AB}=\frac{1}{2}\left(\frac{7R}{3}\right)=\frac{7R}{6}$
Hence option B is correct.