Symmetric Resistor Network

A balanced Wheatstone bridge carries no current through its central galvanometer arm, so the bridging resistor can be removed without altering the current distribution or the equivalent resistance between the supply corners. With the four square arms each equal to 2 (\Omega), the balance condition (P/Q = R/S) is automatically satisfied since all ratios equal one. Ignoring the idle middle resistor, the network reduces to two series pairs in parallel: each path from one supply corner to the opposite corner is (2 + 2 = 4;\Omega), and the two such paths in parallel give (\frac{4 \times 4}{4 + 4} = \frac{16}{8} = 2;\Omega). The value 1 (\Omega) wrongly includes the bridge resistor as a current-carrying parallel element. The value 4 (\Omega) takes only a single series path. The value 5 (\Omega) adds all five resistors in series. This is the classic JEE balanced-bridge simplification. A plausibility check confirms it: the equivalent resistance must lie between the single arm (2 (\Omega)) and the full series path (4 (\Omega)), and the symmetric balanced result of exactly 2 (\Omega) is consistent with the no-current condition in the central arm.