Let $V_1$ be the voltage at a point between the two leftmost resistors in the figure. There are only two resistors that are on horizontal line segments. Let $V_2$ the voltage at a point between them. Let $V_3$ be the voltage at a point between two rightmost resistors in the figure. Then, using Kirchoff's junction rule, we have the following equations
$\begin{align*}\frac{V - V_1}{R} - \frac{V_1}{R} - \frac{V_1 - V_2}{R} &= 0 \\\frac{V - V_3}{R} - \frac{V_3}{R} - \frac{V_...$
When we multiply both sides of these equations by $R$ and then simplify, we obtain
$\begin{align}V - 3 V_1 + V_2 &= 0 \label{eqn:1}\\V - 3 V_3 + V_2 &= 0 \label{eqn:2}\\V - 4 V_2 + V_1 + V_3 &= 0 ...$
The first two equations, when subtracted, imply that $V_1 = V_3$ . Plugging this into equation (3) gives
$\setcounter{equation}{3}\begin{align}V - 4 V_2 + 2 V_1 = 0 \label{eqn:4}\end{align}$
We can plug in $V_2 = 3 V_1 - V$ (from equation (1)) to solve for $V_1$
$\begin{align*}V_1 = \frac{V}{2}\end{align*}$
and then plug this back into equation (4) to get $V_2$
$\begin{align*}V_2 = \frac{V}{2}\end{align*}$
So,
$\begin{align*}V_1 = V_2 = V_3 = \frac{V}{2}\end{align*}$
and the current through the battery is
$\begin{align*}\frac{V_1}{R} + \frac{V_2}{R} + \frac{V_3}{R} = 3 \cdot \frac{V}{2 R} = \boxed{ \frac{3}{2} \frac{V}{R}}\end{al...$
Therefore, answer (D) is correct.

Solution to 2008 Problem 68