Equivalent resistance between A and B in Fig. is,

A. $8r/7$
B. $7r/8$
C. $3r/4$
D. $r$

Hint: First, in order to find the equivalent resistance, the circuit, a network analysis has to be
done and Kirchoff’s current law has to be applied. The current passing through each resistor has
to be identified and based on that the equivalent resistance can be calculated.
Given, the resistance in each resistor is r.
Let us first connect the battery from node A and B in the circuit as shown in Fig.1,

Let voltage from battery is V and the total current flowing is I. A, B, C, D, E, F, G be the nodes
in the circuit. At node A, the current I get divided as shown in Fig.1.
The equivalent resistance is given as,
${R_{eq}} = V/I$ ….…(1)

In the current path, ACEA applying KCL,
$\begin{array}{l}{I_1}r = \dfrac{{I - 2{I_1}}}{2}r + (I - 2{I_1})r\\ \Rightarrow 3I =
8{I_1}\end{array}$ ….…(2)
In the path, FACBGF,
$V = {I_1}r + \dfrac{I}{2}r$ ….…(3)
Substituting the value of I 1 from equation (2) in equation (3),
$\begin{array}{l}V = \dfrac{3}{8}Ir + \dfrac{I}{2}r = \dfrac{7}{8}Ir\\ \Rightarrow
\dfrac{V}{I} = \dfrac{7}{8}r\end{array}$ ….…(4)
Substituting value of V/I in equation (1),
${R_{eq}} = \dfrac{V}{I} = \dfrac{7}{8}r$
Hence, the correct answer is (B).
Note: In the solution, the students can calculate equivalent resistance by initially drawing the
circuit diagram and draw the current flow in each path by connecting a source to the terminals A
and B. The current at each branch or through each resistor has to be obtained. By applying
Kirchoff’s current law, the equation in terms of current and by using Kirchoff’s voltage law the
equation in terms of voltage can be obtained. Once the individual current through each path is
obtained the net current can be obtained and from this the equivalent resistance is obtained.