Current Electricity · MCQ 101 & 102

Equivalent Resistance Between A and B

Two MCQs on bridge-type resistor networks (every resistor = 3 Ω).

101 Single bridge with bottom path
A B 3Ω 3Ω 3Ω 3Ω 3Ω 3Ω 3Ω 3Ω
Top: A→L→(diamond)→R→B. Bottom: A→C→B. All resistors 3 Ω.
Solution — Q101
STEP 1 — Reduce the diamond (between L and R)
Upper route L–P–R = \(3+3 = 6\,\Omega\); lower route L–M–R = \(3+3 = 6\,\Omega\). In parallel: \[ R_{LR} = \frac{6\times6}{6+6} = 3\ \Omega \]
STEP 2 — Top path A→B (series)
\( 3\,(A\text{–}L) + 3\,(L\text{–}R) + 3\,(R\text{–}B) = 9\ \Omega \)
STEP 3 — Bottom path A→B (series)
\( 3\,(A\text{–}C) + 3\,(C\text{–}B) = 6\ \Omega \)
STEP 4 — The two paths are in parallel
\[ R_{AB} = \frac{9\times6}{9+6} = \frac{54}{15} = 3.6\ \Omega \]
(a) 2 Ω (b) 18 Ω (c) 6 Ω (d) 3.6 Ω ✓
Q101 Answer RAB = 3.6 Ω  → (d)
102 Chain of balanced bridges
Heads-up: in your photo the lower diagram and the full option list for Q102 are partly cut off, so treat the count below as the standard form of this problem — verify the number of diamonds against your printed figure.
Key Idea — Balanced Wheatstone Bridge

Each diamond is a balanced bridge (all arms equal, so the ratio \( \tfrac{3}{3} = \tfrac{3}{3} \) is satisfied). In a balanced bridge the diagonal/cross resistor carries zero current and can be deleted.

Each diamond then becomes two equal arms in parallel: \( (3+3)\parallel(3+3) = 3\ \Omega \).

Solution — Q102
STEP 1 — Delete every balanced cross resistor (no current)
The symmetric diamonds all satisfy the balance condition, so each cross resistor is removed.
STEP 2 — Each remaining diamond = 3 Ω, and they sit in series along A→B
With n diamonds in the chain: \( R_{AB} = n \times 3\ \Omega \).
STEP 3 — For the standard 6-section chain shown
\( R_{AB} = 6 \times 3 = 18\ \Omega \).
(a) 54 Ω (b) 18 Ω ✓
Q102 Answer (standard form) RAB = 18 Ω  → (b)
Short Cut (both)

Q101 — two-path trick: a symmetric diamond between two nodes is just (2R)∥(2R) = R. Then the whole thing is one top path ∥ one bottom path: 9 ∥ 6 = 3.6 Ω.

Q102 — balance first: whenever you see equal-arm diamonds, kill the diagonal resistors (zero current) before doing any arithmetic. Each diamond collapses to 3 Ω, and a chain of them simply adds in series: n × 3 Ω.

Q101 verified by nodal analysis (3.6 Ω) · Q102 via balanced-bridge reasoning