Consider the following statements regarding the network shown in the…

The correct answer is Option (4) → (i), (ii) and (iii)

From the figure, the upper branch has \( R \) and \( R \) in series. The lower branch has \( 2R \) and \( 2R \) in series. The resistor \( R' \) joins the middle points of the two branches. The network is a balanced Wheatstone bridge. The ratio of the two resistances is the same in both branches, so the two ends of \( R' \) are at the same potential.

With a potential difference \( V \) between A and B, the fraction of \( V \) dropped across the first resistor of each branch is:

    \( \begin{aligned} \frac{V_{\mathrm{upper}}}{V} &= \frac{R}{R+R} \\ &= \frac{1}{2} \end{aligned} \)

    \( \begin{aligned} \frac{V_{\mathrm{lower}}}{V} &= \frac{2R}{2R+2R} \\ &= \frac{1}{2} \end{aligned} \)

Both middle points are therefore at the same potential, and no current is driven through \( R' \).

(i) The equivalent resistance of the network between A and B is independent of the value of \( R' \). — True. No current flows through \( R' \) whatever its value, so it can be removed without changing any current in the network.

(ii) The equivalent resistance of the network between A and B is \( 4R/3 \). — True. With \( R' \) removed, the upper branch \( 2R \) is in parallel with the lower branch \( 4R \):

    \( \begin{aligned} R_{AB} &= \frac{R_{\mathrm{upper}}\,R_{\mathrm{lower}}}{R_{\mathrm{upper}}+R_{\mathrm{lower}}} \\ &= \frac{2R\times4R}{2R+4R} \\ &= \frac{4R}{3} \end{aligned} \)

(iii) The current flowing through \( R' \) is zero. — True. The two ends of \( R' \) are at the same potential, so there is no potential difference across it and no current through it.

NCERT (2026-27):
Wheatstone Bridge – “… Of special interest, is the case of a balanced bridge where the resistors are such that I_g = 0.”