In the given circuit, the equivalent resistance across the 3 V batter…

The correct answer is Option (3) → \( 75\,\Omega \)

The network is reduced step by step, starting from the part farthest from the battery. Resistors in series add, \( R_s=R_1+R_2 \). Two resistors in parallel give \( R_p=R_1R_2/(R_1+R_2) \).

From the figure, \( R_3 \) and \( R_4 \) are both joined between the upper wire and the top of \( R_5 \). So they are in parallel:

    \( \begin{aligned} R_{34} &= \frac{R_3R_4}{R_3+R_4} \\ &= \frac{60\times30}{60+30}\,\Omega \\ &= 20\,\Omega \end{aligned} \)

This pair is in series with \( R_5 \):

    \( \begin{aligned} R' &= R_{34}+R_5 \\ &= 20\,\Omega+30\,\Omega \\ &= 50\,\Omega \end{aligned} \)

This branch and \( R_2 \) both connect the upper and lower wires, so they are in parallel:

    \( \begin{aligned} R'' &= \frac{R'R_2}{R'+R_2} \\ &= \frac{50\times50}{50+50}\,\Omega \\ &= 25\,\Omega \end{aligned} \)

\( R_1 \) carries the whole current from the battery, so it is in series with the rest:

    \( \begin{aligned} R_{\mathrm{eq}} &= R_1+R'' \\ &= 50\,\Omega+25\,\Omega \\ &= 75\,\Omega \end{aligned} \)