Each resistance in the given cubical network has resistance of $1 \Omega$ and equivalent resistance between $A$ and $B$ is

• Correct answer
  A

  $\frac{5}{6} \Omega$
• B

  $\frac{6}{5} \Omega$
• C

  $\frac{5}{12} \Omega$
• D

  $\frac{12}{5} \Omega$

Solution

The given network of resistances can be shown as follows,

Let $6 I$ be the current through the cell, which is distributed as shown in the figure. Applying Kirchhoff's law to the loop $A B C C^{\prime} E A$, we have $\therefore \quad E=5 I$ If $R^{\prime}$ be the equivalent resistance of the network, then $R^{\prime}=\frac{\text { total emf }}{\text { total current }}=\frac{5 I}{6 I}=\frac{5}{6} \Omega$

Hence, the correct answer is option: 'A'.