Question

In the circuit shown in figure, what will be the effective resistance between point $A$ and $B$ ?

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

Answer 1

The correct option is B $\frac{r}{2}$

The upper five resistance between $A$ and $B$ will form a balanced Wheatstone bridge hence there will be no current in the arm $CO$

the equilibrium circuit will be shown in fig.$2$

hence the resistance of arm $ACB,AOB,$ and $AB$ are all in parallel

the effective resistance between $A$ and $B$ is given as

$\begin{array}{c}\frac{1}{R}=\frac{1}{\left(r+r\right)}+\frac{1}{\left(r+r\right)}+\frac{1}{r}\\ =\frac{1}{2r}+\frac{1}{2r}+\frac{1}{r}\\ =\frac{2}{2r}+\frac{1}{r}\\ =\frac{1}{r}+\frac{1}{r}\\ \Rightarrow \frac{1}{R}=\frac{2}{r}\\ \Rightarrow R=\frac{r}{2}\Omega \end{array}$