--- title: "Source current by series-parallel reduction" subject: "Basic Electrical Engineering" chapter: 2 key: "semester_1/basic_electrical_engineering:series-parallel-current-30v-r1-r9" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/series-parallel-current-30v-r1-r9 site: BCT Engineering --- # Source current by series-parallel reduction Basic Electrical Engineering · Chapter 2: DC circuits - Asked in: 2069 Asar, 6 marks - Video: [Resistors In Series and Parallel Circuits - Keeping It Simple!](https://www.youtube.com/watch?v=dEuBykX5_B4) — The Organic Chemistry Tutor - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/series-parallel-current-30v-r1-r9 - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2 ## Question 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.] ## 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. ## Answer #### 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** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. R1 is directly across the source, so A. it is in series with R2 B. it carries no current C. it is shorted D. it is in parallel with the rest of the network Answer: D. The source current is the sum of R1's current and the rest. (Key point: R1 position) 2. R2 + R3 = 4 Ω in parallel with R9 = 4 Ω gives A. 8 Ω B. 4 Ω C. 2 Ω D. 1 Ω Answer: C. 4||4 = 2 Ω. (Key point: A to C) 3. R8 = 2 Ω in parallel with (R4 + R5 = 3 Ω) gives A. 5 Ω B. 1.2 Ω C. 1 Ω D. 2.5 Ω Answer: B. 2 x 3/5 = 1.2 Ω. (Key point: C to D) 4. R_total is 1.0545 Ω, so the source current is A. 30/1.0545 = 28.45 A B. 15 A C. 13.45 A D. 30 A Answer: A. R1 takes 15 A and the rest 13.45 A, adding to 28.45 A. (Key point: Total) ## Related in Basic Electrical Engineering - [Series-parallel AC circuit analysis](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-parallel-ac-r1-15-l1-0-05-200v) — Basic Electrical Engineering, chapter 6, asked 2× - [Current from source in ladder circuit](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Asource-current-3v-2ohm-ladder) — Basic Electrical Engineering, chapter 2 - [Series-parallel circuit with I1, I2, I3 phasors](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-parallel-ac-v100-z1-z2-z3) — Basic Electrical Engineering, chapter 6, asked 2× - [Series-parallel circuit currents and power](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-parallel-z1-6-0-01h-100v) — Basic Electrical Engineering, chapter 6 - [Source current from parallel branch currents](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aparallel-branch-currents-time-expression) — Basic Electrical Engineering, chapter 5 - [Series versus parallel circuit comparison](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-parallel-comparison-table) — Basic Electrical Engineering, chapter 1 ## Related in other subjects - [Show i1 = i2 = -V/(R1+R2+R3) at t=0+](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aswitch-a-to-b-initial-values) — Electrical Circuit Theory, chapter 2 - [Nodal analysis current in 6 Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Anodal-analysis-6-ohm-current) — Basic Electrical and Electronics Engineering, chapter 1 - [Series-parallel AC circuit currents and power](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Aseries-parallel-ac-circuit-power) — Fundamental of Electrical and Electronics Engineering, chapter 3 - [Power factor and series-parallel circuit calculation](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Apower-factor-series-parallel-circuit) — Fundamental of Electrical and Electronics Engineering, chapter 3