--- title: "Kirchhoff law branch currents" subject: "Basic Electrical Engineering" chapter: 2 key: "semester_1/basic_electrical_engineering:kirchhoff-law-i1-i2-i3-12v-20v" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/kirchhoff-law-i1-i2-i3-12v-20v site: BCT Engineering --- # Kirchhoff law branch currents Basic Electrical Engineering · Chapter 2: DC circuits - Asked in: 2071 Shrawan, 6 marks - Video: [How to Solve a Kirchhoff's Rules Problem - Simple Example](https://www.youtube.com/watch?v=Z2QDXjG2ynU) — Jesse Mason - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/kirchhoff-law-i1-i2-i3-12v-20v - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2 ## Question Find I1, I2 and I3 in the circuit shown in the figure using Kirchhoff's law. [Figure: 12 V source on the left with 4 Ω on top and 3 Ω below; central 5 Ω resistor; 6 Ω on the top right with a 20 V source; 7 Ω on the right; bottom branch has 8 Ω with a 10 V source.] ## Short answer Apply KVL: loop 1 gives 3I1 - 7I2 - 5I3 = -12, loop 2 gives 7I1 - 13I2 + 18I3 = -20, loop 3 gives 18I1 - 10I2 + 7I3 = 10. Solving yields I1 = 2.00 A, I2 = 2.59 A, I3 ≈ -0.02 A. ## Answer #### Reading the figure Nodes: Lw (left middle), P (top middle), M (centre) and Rw (right middle). - Lw to P: 12 V battery (+ up) then 4 Ω; current I₂ flows up from Lw to P. - P to M: 5 Ω, current I₃ downward. - P to Rw: 6 Ω then 20 V battery (+ at the Rw end). - Lw to M: 3 Ω. M to Rw: 7 Ω. - Rw to Lw: 10 V battery (+ on the 8 Ω side) then 8 Ω; current I₁ flows up into Lw. Polarities are taken from the plate lengths in the scan. #### Currents by KCL - Lw: I₁ enters, I₂ leaves, so the 3 Ω carries I₁ − I₂ (Lw to M). - M: 5 Ω brings I₃, 3 Ω brings I₁ − I₂, so the 7 Ω carries I₁ − I₂ + I₃ (M to Rw). - P: I₂ enters, I₃ leaves, so the 6 Ω carries I₂ − I₃ (P to Rw). #### Loop equations by KVL Loop 1 (Lw, 12 V, 4 Ω, 5 Ω, 3 Ω): 12 − 4 I₂ − 5 I₃ + 3 (I₁ − I₂) = 0 ⇒ 3 I₁ − 7 I₂ − 5 I₃ = −12 Loop 2 (P, 6 Ω, 20 V, 7 Ω, 5 Ω): −6 (I₂ − I₃) + 20 + 7 (I₁ − I₂ + I₃) + 5 I₃ = 0 ⇒ 7 I₁ − 13 I₂ + 18 I₃ = −20 Loop 3 (Lw, 3 Ω, 7 Ω, 10 V, 8 Ω): −3 (I₁ − I₂) − 7 (I₁ − I₂ + I₃) + 10 − 8 I₁ = 0 ⇒ 18 I₁ − 10 I₂ + 7 I₃ = 10 #### Solution I₁ = 2.00 A, I₂ = 2.59 A, I₃ = −0.021 A The negative sign means I₃ actually flows upward (from M to P), but it is very small. **Answer: I₁ ≈ 2.00 A, I₂ ≈ 2.59 A, I₃ ≈ −0.02 A (about zero)** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At node Lw, I1 enters and I2 leaves, so the 3 Ω carries A. I1 + I2 B. I2 - I1 + I3 C. I1 D. I1 - I2 Answer: D. KCL: the remainder of I1 after I2 leaves goes through the 3 Ω. (Key point: KCL at Lw) 2. The 7 Ω carries A. I1 - I2 - I3 B. I1 + I2 + I3 C. I1 - I2 + I3 D. I3 Answer: C. At M the 3 Ω brings I1 - I2 and the 5 Ω brings I3, so both flow to the 7 Ω. (Key point: KCL at M) 3. Loop 1 equation is A. 3 I1 + 7 I2 + 5 I3 = 12 B. 3 I1 - 7 I2 - 5 I3 = -12 C. I1 - I2 = -12 D. 7 I1 - 13 I2 + 18 I3 = -20 Answer: B. 12 - 4 I2 - 5 I3 + 3(I1 - I2) = 0 simplifies to this. (Key point: Loop 1) 4. I3 = -0.021 A means A. I3 actually flows opposite to the assumed direction, and it is almost zero B. the 5 Ω has negative resistance C. I3 is large D. the solution is invalid Answer: A. A negative sign only indicates direction; it is the opposite of assumed. 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