Question

Thirteen resistance each of resistance $R$ ohm are connected in the circuit as shown in the figure. The effective resistance between $A$ and $B$ is

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

Answer 1

The correct option is D $\frac{2R}{3}\Omega$

In given configuration for the part $AC,R_1$ and $R_2$ are in series and their resultant is in parallel with the $R_3$

$⟹\frac{1}{R_{eqAC}}=\frac{1}{R}+\frac{1}{2R}$

$⟹R_{eqAC}=\frac{2}{3}R$

Similarly for part $AD,CB,DB$

$⟹R_{eqAD}=\frac{2}{3}R,⟹R_{eqDB}=\frac{2}{3}R$

$⟹R_{eqCB}=\frac{2}{3}R$

$\therefore$ due to equal resistance in part $AC$ and $AD$ there will be no potential difference across arm $CD$ and hence that arm will not contribute in resultant resistor.

$\therefore$ Also $⟹$ that the arm $AC$ is in series with $CB$ and their resultant is in parallel with the series combination of $AD$ and $DB$.

$⟹R_1=R_{AC}+R_{AB}=\frac{2R}{3}+\frac{2R}{3}=\frac{4R}{3}$

Also, $R_1=R_{AD}+R_{DB}=\frac{2R}{3}+\frac{2R}{3}=\frac{4R}{3}$

$⟹\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R_{eq}}=\frac{3}{4R}+\frac{3}{4R}=\frac{6}{4R}=\frac{3}{2R}$

$⟹R_{eq}=\frac{2R}{3}$