Chapter 1

Q. 1.P.12

Twelve 1 Ω resistors are used as edges to form a cube. The R_{ eq }  seen between the two diagonally opposite corners of cube is

(a) \frac{6}{5} Ω       (b) \frac{5}{6} Ω       (c) 6 Ω       (d) \frac{5}{12} Ω



Verified Solution

The equivalent circuits is

Applying KVL, we get

\begin{gathered} V-\frac{I}{3} R-\frac{I}{6} R-\frac{I}{3} R=0 \\ V=I\left\lgroup \frac{R}{3}+\frac{R}{6}+\frac{R}{3} \right\rgroup=I \frac{5 R}{6} \\ \Rightarrow \frac{V}{I}=R_{ eq }=\frac{5}{6}  \Omega \end{gathered}