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Source current by series-parallel reduction

Short answer

Combine resistances: R2+R3 in parallel with R9 gives 2 Ω, then with R8 in parallel with (R4+R5) yields 1.2 Ω, combine with R7 to 1.231 Ω, add R6 gives 2.231 Ω, parallel with R1 gives 1.0545 Ω. I = 30 / 1.0545 ≈ 28.4 A.

Using series-parallel combination of resistances find the current delivered by the source in the following circuit. [Figure: 30 V source; points A, B, C, D, E; R1 = 2 Ω from B to A; R2 = 2 Ω from A to C; R3 = 2 Ω from C towards E; R9 = 4 Ω from A to C; R7 = 2 Ω from A to D; R8 = 2 Ω from C to D; R4 = 2 Ω from C to E; R6 = 1 Ω from B to D; R5 = 1 Ω from D to E.]

Reading the figure

Nodes A, B, C, D, E and the apex K. The 30 V source (+ on top) is connected across A and B (its + wire goes to A, its − wire to B).

Branches: R1 = 2 Ω (A–B), R2 = 2 Ω (A–K) and R3 = 2 Ω (K–C), R9 = 4 Ω (A–C), R7 = 2 Ω (A–D), R8 = 2 Ω (C–D), R4 = 2 Ω (C–E), R5 = 1 Ω (D–E), R6 = 1 Ω (B–D).

R1 is directly across the source. The current from the source is the sum of the currents in R1 and in the rest of the network.

Step 1: Upper part A to C

R2 + R3 = 4 Ω in parallel with R9 = 4 Ω gives 2 Ω.

Step 2: Between C and D

R8 = 2 Ω in parallel with (R4 + R5 = 3 Ω): 2 × 3/5 = 1.2 Ω

Step 3: Between A and D

Path A–C–D = 2 + 1.2 = 3.2 Ω in parallel with R7 = 2 Ω: 2 × 3.2/5.2 = 1.231 Ω

Step 4: Add R6

Path A–D–B = 1.231 + 1 = 2.231 Ω

Step 5: Parallel with R1

R_total = 2 × 2.231/(2 + 2.231) = 1.0545 Ω

Source current

I = V/R_total = 30/1.0545 = 28.45 A

(Check: R1 carries 15 A and the rest 13.45 A.)

Answer: Source current ≈ 28.4 A

✓ Answer checked by Claude Opus 5.5 (high effort)

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