Current electricityCard 12 · NR seriesNEET 2027

Resistance-network reduction

Five Tier-1 practice items — the three network shapes NEET returns to most often: bridge balance, bent uniform wire, and redraw-by-node-labelling.

5 items Target 90 s each Cell resistance ignored unless stated

NR-01Bridge balance → equivalent resistance ▮▮▮▮▯

In the network shown, find the equivalent resistance between terminals A and C.

Fig. NR-01

a5 Ω
b3.6 Ω
c10/3 Ω
d15 Ω

(b) 3.6 Ω

Check the ratio before computing anything: 2/4 = 3/6, so the bridge is balanced. B and D sit at the same potential, no current crosses the 10 Ω arm, and it can be deleted. What remains is (2 + 4) in parallel with (3 + 6) = 6 ∥ 9 = 3.6 Ω.

Trap → 10/3 Ω comes from shorting B to D instead of opening the arm. For a balanced bridge both moves give the same result, so 10/3 only appears if the ratio was never checked in the first place.

NR-02Bent uniform wire → arcs in parallel ▮▮▮▮▮

A uniform wire of total resistance 12 Ω is bent into a complete circle. Points A and B on the rim subtend 90° at the centre. The resistance between A and B is:

Fig. NR-02

a3.00 Ω
b2.40 Ω
c2.25 Ω
d9.00 Ω

Resistance of a uniform wire is proportional to its length, so the arcs split in the ratio 90 : 270, giving 3 Ω and 9 Ω. The two arcs are the only two paths from A to B, so they are in parallel: (3 × 9)/12 = 2.25 Ω.

Trap → options (a) and (d) are the two arcs taken alone. Both are picked by students who find the arcs correctly and then stop before combining them.

NR-03Shorting wire → node merge ▮▮▮▮▯

Four resistors form a square ABCD as shown. The diagonal BD is a connecting wire of negligible resistance. The resistance between A and C is:

Fig. NR-03

a2 Ω
b3 Ω
c5 Ω
d7 Ω

The wire makes B and D a single node — relabel both as one point M. From A there are now two ways to M: 3 Ω and 6 Ω, in parallel = 2 Ω. From M to C: 4 Ω and 12 Ω, in parallel = 3 Ω. These two blocks are in series, so 2 + 3 = 5 Ω.

Trap → 7 Ω is the A–B–C path taken alone. Removing the diagonal entirely would instead give 7 ∥ 18 = 5.04 Ω — close enough to 5 that a student who ignores the wire may still feel right, which is exactly why the sides are unequal here.

NR-04Bridge balance → branch current ▮▮▮▮▯

A cell of emf 10 V and negligible internal resistance is connected across the network below. The current through the 25 Ω arm is:

Fig. NR-04

a0.50 A
b0.25 A
c0.20 A
dzero

(d) zero

10/20 = 20/40, so the bridge is balanced and B and D are at the same potential. With no potential difference across it, the 25 Ω arm carries no current — whatever the emf. The arm's value is irrelevant to the answer, which is the point of the question.

Trap → 0.50 A is the total current drawn from the cell (R_eq = 30 ∥ 60 = 20 Ω). It is the most-picked wrong option because the student does the reduction correctly and then answers a question that wasn't asked.

NR-05Node labelling → hidden parallel bank ▮▮▮▯▯

In the arrangement below, the two horizontal rails are connecting wires of negligible resistance. The resistance between A and B is:

Fig. NR-05

a24 Ω
b12 Ω
c4/3 Ω
d1 Ω

(d) 1 Ω

A rail of zero resistance is a single node however long it is drawn. Every resistor therefore has its top end at A and its bottom end at B — all four are in parallel. 1/2 + 1/4 + 1/6 + 1/12 = 1, so R = 1 Ω.

Trap → 24 Ω is the series sum. The ladder shape is what suggests series; the give-away that it isn't is that both terminals hang off the same end of the structure.

Answer key

ID	Archetype	Answer	Move that solves it
NR-01	Bridge balance	3.6 Ω	Check P/Q = R/S first, then delete the arm
NR-02	Bent uniform wire	2.25 Ω	Split by arc length, combine the two arcs in parallel
NR-03	Shorting wire	5 Ω	Merge the shorted pair into one node, then relabel
NR-04	Bridge balance	zero	Balanced ⇒ no potential difference across the arm
NR-05	Node labelling	1 Ω	Zero-resistance rail = one node; read the endpoints

NR series · card 12 of the Current Electricity set · prepared for NEET 2027