A battery of emf 12 V and negligible resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance $2 \Omega$. The equivalent resistance of the network is

• A

  $\frac{1}{2} \Omega$
• B

  $\frac{4}{3} \Omega$
• C

  $\frac{1}{4} \Omega$
• Correct answer
  D

  $\frac{5}{3} \Omega$

Solution

In a closed loop, say $A B C C^{\prime} E A$ and apply Kirchhoffs second rule, $-I R-\left(\frac{I}{2}\right) R-I R+\varepsilon=0$

where, $R$ is the resistance of each edge and $\varepsilon$ the emf of battery.
Thus, $\varepsilon=\frac{5}{2} I R$
The equivalent resistance $R_{\text {eq }}$ of the network is
$R_{\mathrm{eq}}=\frac{\varepsilon}{3 I}=\frac{5}{6} R=\frac{5}{6} \times 2=\frac{5}{3} \Omega$

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