Current Electricity · MCQ 60

Equivalent Resistance Between A and B

Five 1 Ω resistors lie in a row from A to B, with one wire bridging the top and one bridging the bottom. Find R_AB (in Ω).

A–P–Q–R–S–B (each gap = 1 Ω). Top wire joins Q–S; bottom wire joins P–R.

⬇ merge the wire-shorted nodes

A–X = 1 Ω · three 1 Ω in parallel between X and Y · Y–B = 1 Ω.

Given

Five resistors in a row, each 1 Ω: A–P, P–Q, Q–R, R–S, S–B.
Top jumper (no resistor): Q to S.
Bottom jumper (no resistor): P to R.

Asked

Equivalent resistance R_AB.

Key Idea + Formula

Plain wire ⇒ both ends are one node. Then series adds; parallel:

\[ \frac{1}{R_{\parallel}} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} \]

Solution

STEP 1 — Merge the jumpered nodes

Bottom wire ⇒ P ≡ R (call it X). Top wire ⇒ Q ≡ S (call it Y).

STEP 2 — Re-read each resistor by its end nodes

Resistor	Ends	After merge
A–P	A, P	A – X
P–Q	P, Q	X – Y
Q–R	Q, R	Y – X
R–S	R, S	X – Y
S–B	S, B	Y – B

Three resistors (P–Q, Q–R, R–S) all bridge X to Y.

STEP 3 — Combine

Middle bundle: \( 1\parallel1\parallel1 = \tfrac{1}{3}\,\Omega \). Then series with A–X and Y–B: \[ R_{AB} = 1 + \frac{1}{3} + 1 = \frac{7}{3}\ \Omega \]

(a) 1/5 (b) 5/4 (c) 7/3 ✓ (d) 7/2

Answer R_AB = 7⁄3 Ω → (c)

Short Cut

Letter the nodes, merge the wires, group by end-pair. The two jumpers fold the middle three resistors onto the same node pair (X, Y), so they instantly become 1∥1∥1 = 1/3 Ω. The first and last resistors are alone in series: 1 + 1/3 + 1 = 7/3 Ω.

Pattern: staggered jumpers like these always sandwich the inner resistors into one parallel block — count how many fall between the two merged nodes, that's your 1/n.

Given → Asked → Formula → Solution → Short Cut · verified by nodal analysis (7/3 Ω)