Question

In the arrangement shown, the magnitude of each resistance is $2 \,\Omega$. The equivalent resistance between $O$ and $A$ is given by

• A. $\frac{14}{15} \Omega$
• B. $\frac{7}{15} \Omega$
• C. $\frac{4}{3} \Omega$
• D. $\frac{5}{6} \Omega$

Answer 1

Answer: (A)

Since magnitude of each resistance $2 \Omega$ then points $B$ and $D$ are equipotential.
$\Rightarrow V_{B}-V_{D}=0$
Therefore we can connect these points with a conducting wire.

$R_{e q}=\left(\frac{3}{4}+1\right) \| 2$
$=\frac{\frac{7}{4} \times 2}{\frac{7}{4}+2}=\frac{\frac{7 \times 2}{4}}{\frac{7+8}{4}}$
$R_{e q}=\frac{14}{15} \Omega$