In the network of resistances shown below, Current through the battery will be ___________.

Question

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Correct — option (4), 1.0 A. Label the node at the battery's negative terminal A and the node at its positive terminal D. Tracing the printed diagram gives two junctions between A and D: node Q, where the 1-ohm resistor (from A) meets the second 2-ohm resistor (leading to D), and node B, where the first 2-ohm resistor (from A) meets the 4-ohm resistor (leading to D); the 3-ohm resistor connects B and Q directly, bridging the two branches. This is a Wheatstone-bridge layout, and it happens to be balanced: a bridge is balanced when the ratio of the two resistors in one arm equals the ratio of the two resistors in the other arm, measured from the same starting node. Here, going through B, the ratio is (A-to-B)/(B-to-D) = 2/4 = 1/2; going through Q, the ratio is (A-to-Q)/(Q-to-D) = 1/2 = 1/2. The two ratios are equal, so the bridge is balanced and, by the balanced-bridge property, no current flows through the bridging 3-ohm resistor at all — both its ends sit at the same potential. With the 3-ohm resistor carrying no current, it can be removed from the circuit for the purpose of calculating total current, leaving two independent parallel paths from A to D: the path through B, with resistance 2 + 4 = 6 ohms, and the path through Q, with resistance 1 + 2 = 3 ohms. Their parallel combination is (6 x 3)/(6 + 3) = 18/9 = 2 ohms, which is the network's total equivalent resistance across the battery. By Ohm's law, the current drawn from the 2 V battery is I = V/R = 2/2 = 1.0 A, matching option (4) exactly.

Why the other options are wrong — (a) 1.5 A would require a total equivalent resistance of 2/1.5 = 1.33 ohms, which does not match the network's actual 2-ohm equivalent resistance. A value like this typically comes from treating the bridge as unbalanced and folding the 3-ohm resistor into a series-parallel calculation instead of first checking the balance condition (2/4 = 1/2, matching 1/2) that lets it be dropped.  (b) 3.5 A would require a total resistance of 2/3.5 which is well under 1 ohm, far lower than the network can actually present — even the smaller of the two parallel branches (the 1-ohm-plus-2-ohm path) alone is 3 ohms. This value is consistent with mistakenly summing conductances or resistances in the wrong combination, producing an implausibly low total resistance.  (c) 4.0 A would require a total resistance of only 0.5 ohm, smaller than any single resistor in the network and smaller than either full branch (3 ohms and 6 ohms) computed independently. A resistor network built purely from series and parallel combinations of 1, 2, 2, 3 and 4-ohm resistors cannot present less resistance than its smallest single resistor (1 ohm) in this configuration, so this value does not correspond to any correct reading of the circuit.

Concept: A Wheatstone bridge is a four-resistor network arranged as two parallel branches between a common source and a common return, each branch made of two resistors in series, with a fifth 'bridge' resistor (or galvanometer, in the classic measuring instrument) connecting the midpoint of one branch to the midpoint of the other. The bridge is balanced exactly when the ratio of the two resistors in one branch equals the ratio of the two resistors in the other branch (R1/R2 = R3/R4, read from the same end); when that condition holds, both midpoints sit at the same electric potential, so no current flows through the bridge resistor regardless of its own value, and it can be deleted from the circuit for the purpose of computing total resistance or current, leaving a simple parallel combination of the two branches.

State PCS 2025 Q52 — Science & Technology PYQ with Answer

State PCS 2025 Q52 — Science & Technology PYQ with Answer