Current Electricity · Resistor Network

Equivalent Resistance Between A and E

Find the equivalent resistance between the ends A and E of the network shown — a triangle BCD with two tail resistors leading out to A and E.

Continuation note. Same Current Electricity chapter as the previous examples. The book's own line confirms the plan: "resistance R₁, R, R₄ form a series combination," where R is the equivalent of the B–D triangle portion. No numerical values are supplied, so the answer is symbolic.

R₂ = B–C, R₃ = C–D, R₅ = B–D, R₁ = B–A tail, R₄ = D–E tail.

Given

R₂ between B and C, R₃ between C and D (upper path).
R₅ directly between B and D (lower path).
R₁ tail from B to terminal A.
R₄ tail from D to terminal E.

Asked

Equivalent resistance R_AE between the two outer ends A and E.

Formula

Series: R = R₁ + R₂ + … | Parallel (two arms):

\[ R_{\parallel} = \frac{R_a R_b}{R_a + R_b} \]

Solution

STEP 1 — The two B→D paths are in series within each arm

Upper arm through C: R₂ and R₃ in series → \( R_2 + R_3 \). Lower arm: just R₅.

STEP 2 — Those two arms are in parallel between B and D

Call this combined resistance R: \[ R = \frac{(R_2 + R_3)\,R_5}{(R_2 + R_3) + R_5} \]

STEP 3 — R₁, R, R₄ are now a simple series chain A → B → D → E

\[ R_{AE} = R_1 + R + R_4 \]

Result \( R_{AE} = R_1 + R_4 + \dfrac{(R_2 + R_3)\,R_5}{R_2 + R_3 + R_5} \)

Short Cut

Spot the parallel "island" first. Only the part between the two junctions B and D involves any parallel work — everything hanging off the outside (R₁ and R₄) is pure series and just gets added on.

So the whole problem reduces to one mental move: (R₂+R₃) ∥ R₅, then + R₁ + R₄. The C-vertex is a dead giveaway — any path through a single intermediate vertex is "the two side resistors in series."

Given → Asked → Formula → Solution → Short Cut · symbolic (no numeric values given)