Current Electricity · Example 3.24

Equivalent Resistance Between A and B

Find the equivalent resistance between A and B in the following two cases.

Source note. This is Example 3.24 from a Current Electricity chapter (Chapter 3) of a standard NEET / JEE physics reference book — the worked-example numbering style (3.23, 3.24…), the preceding "wire bent into a circle ⇒ 3R/2" example, and these network values are characteristic of Indian competitive-exam physics texts (e.g. DC Pandey-style Electricity & Magnetism / Arihant-type compilations). I couldn't verify the exact title from the page image alone, so I'm flagging this as a probable identification rather than a confirmed one. If you can show the chapter cover or contents page, I can pin it down precisely.

i Triangle network (Δ)

A–C = 20 Ω, C–B = 30 Ω, A–B = 50 Ω. Find R across A–B.

Given

Side A–C: 20 Ω
Side C–B: 30 Ω
Side A–B: 50 Ω

Asked

Equivalent resistance R_AB between terminals A and B.

Formula

Series: R = R₁ + R₂. Parallel:

\[ R_{\text{eq}} = \frac{R_a\,R_b}{R_a + R_b} \]

Solution

STEP 1 — Identify the two paths from A to B

Path 1 (direct): the 50 Ω side.
Path 2 (via C): 20 Ω and 30 Ω are in series → \( 20 + 30 = 50\ \Omega \).

STEP 2 — The two paths share the same ends (A and B) ⇒ parallel

\[ R_{AB} = \frac{50 \times 50}{50 + 50} = \frac{2500}{100} = 25\ \Omega \]

Case (i) R_AB = 25 Ω

Short Cut

When two paths between the same nodes happen to be equal (50 Ω ∥ 50 Ω), the answer is simply half of one of them → 50 ÷ 2 = 25 Ω. No fraction needed.

ii Ladder / bridge-style network

A→X = 8 Ω, X→Y = 4 Ω, X→bottom = 3 Ω, Y→bottom = 1 Ω. Bottom rail = node B.

Given

A to node X: 8 Ω (top branch)
X to node Y: 4 Ω (top branch)
X down to bottom rail: 3 Ω
Y down to bottom rail: 1 Ω
The whole bottom rail is one node = B.

Asked

Equivalent resistance R_AB between A and B.

Formula

Reduce from the far end inward using series & parallel:

\[ R_{\text{series}} = R_1 + R_2, \qquad R_{\text{parallel}} = \frac{R_a R_b}{R_a + R_b} \]

Solution

STEP 1 — Combine the far branch (right side)

Going X → Y → bottom, the 4 Ω and 1 Ω are in series: \[ R_{4+1} = 4 + 1 = 5\ \Omega \]

STEP 2 — This 5 Ω is in parallel with the 3 Ω

Both the 3 Ω branch and the (4+1) Ω branch connect node X to the bottom rail (B): \[ R_{XB} = \frac{3 \times 5}{3 + 5} = \frac{15}{8} = 1.875\ \Omega \]

STEP 3 — Add the 8 Ω in series (A → X)

\[ R_{AB} = 8 + \frac{15}{8} = \frac{64 + 15}{8} = \frac{79}{8}\ \Omega \]

Case (ii) R_AB = 79⁄8 = 9.875 Ω

Short Cut

Work from the end farthest from the source toward A. The 4 Ω and 1 Ω are "dangling" in series, so collapse them first (→ 5 Ω), then the obvious 3 Ω ∥ 5 Ω, then drop in the lone series 8 Ω last. Reducing outside-in avoids redrawing the circuit.

Trap to avoid: the 8 Ω is not part of any parallel pair — it sits alone between A and X, so it's pure series and just gets added at the very end.

Six-step recall: Given → Asked → Formula → Solution → Short Cut · Source noted above