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.

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