Current Electricity · MCQ 29 & 30

Resistance Between Two Points

Two quick network MCQs: a resistor triangle and a bridge-type diamond.

29 Triangle of three 2 Ω resistors

Measure across any two corners (say v₁ and v₂).

Solution — Q29

Two paths between any two vertices

Direct side = 2 Ω; the other two sides in series = 2 + 2 = 4 Ω.

They are in parallel

\[ R = \frac{2 \times 4}{2 + 4} = \frac{8}{6} = \frac{4}{3}\ \Omega \]

(a) 4/3 Ω ✓ (b) 3/4 Ω (c) 3 Ω (d) 6 Ω

Q29 Answer R = 4⁄3 Ω → (a)

30 Diamond with a 6 Ω diagonal

A–D, D–C, A–B, B–C each 3 Ω; the diagonal A–C is 6 Ω. Find R across A–B.

Solution — Q30

STEP 1 — Reduce A to C

Two equal routes: direct A–C = 6 Ω, and A–D–C = 3 + 3 = 6 Ω. In parallel: \[ R_{AC} = \frac{6 \times 6}{6 + 6} = 3\ \Omega \]

STEP 2 — Continue A→C→B (series)

\( R_{A\to C\to B} = 3\,(A\text{–}C) + 3\,(C\text{–}B) = 6\ \Omega \)

STEP 3 — Parallel with the direct A–B (3 Ω)

\[ R_{AB} = \frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2\ \Omega \]

(a) 5 Ω (b) 2 Ω ✓ (c) 3 Ω (d) 4 Ω

Q30 Answer R_AB = 2 Ω → (b)

Short Cut (both)

Q29 — triangle rule: equal sides R ⇒ across any two corners is R ∥ 2R = 2R/3. With R = 2: 4/3 Ω in one step.

Q30 — collapse the easy node first: D only links A and C, so A–C is two equal 6 Ω paths ⇒ 3 Ω. Then it's a clean series-then-parallel: (3+3) ∥ 3 = 2 Ω. (The diagram is a balanced bridge, so you could also note no current flows where the arms balance.)

Given → Asked → Formula → Solution → Short Cut · both verified numerically