--- title: "Loop current method, current in 1Ω" subject: "Basic Electrical Engineering" chapter: 2 key: "semester_1/basic_electrical_engineering:loop-current-1-ohm-4a-12v-6v" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/loop-current-1-ohm-4a-12v-6v site: BCT Engineering --- # Loop current method, current in 1Ω Basic Electrical Engineering · Chapter 2: DC circuits - Asked in: 2071 Chaitra, 8 marks - Video: [Mesh Analysis or Maxwell’s Loop Current Method | Electrical Engineering](https://www.youtube.com/watch?v=E5M4WvJds_A) — Electrical and Electronics Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/loop-current-1-ohm-4a-12v-6v - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2 ## Question Calculate the current flowing through 1 ohm resistor for network shown below using loop current method. [Figure: 4 A current source on top; 12 V source with 2 Ω on the left; 2 Ω in the upper middle; 1 Ω shunt in the middle; 6 V source on the right.] ## Short answer With clockwise loop currents, I1 = 3.75 A, I2 = −0.75 A. The current through the 1 Ω resistor is I1 − I2 = 4.5 A, flowing downward. ## Answer #### Reading the figure Both batteries have + on top. Left node L (top of 12 V) and right node R (top of 6 V). Between them: 2 Ω from L to the middle node M, 2 Ω from M to R, and 1 Ω from M to the bottom rail. The 4 A current source sits across the top between L and R. The 4 A source is connected between two nodes whose voltages are fixed by the batteries (L = 12 V, R = 6 V), so it does not change the current in the 1 Ω. #### Loop currents (clockwise) - I₁: left loop (12 V, 2 Ω, 1 Ω) - I₂: right loop (1 Ω, 2 Ω, 6 V) Left loop: 12 = 2 I₁ + 1 (I₁ − I₂) ⇒ 3 I₁ − I₂ = 12 Right loop (going down through the 6 V from + to −): 0 = 2 I₂ + 6 + 1 (I₂ − I₁) ⇒ −I₁ + 3 I₂ = −6 #### Solution From the first equation I₂ = 3 I₁ − 12. Then −I₁ + 9 I₁ − 36 = −6, so I₁ = 3.75 A and I₂ = −0.75 A. #### Current in the 1 Ω I = I₁ − I₂ = 3.75 + 0.75 = 4.5 A (downward) **Answer: I₁Ω = 4.5 A** Check by nodal method: V_M = (12/2 + 6/2)/(1/2 + 1/2 + 1) = 9/2 = 4.5 V, so I = 4.5 A. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does the 4 A source not affect the 1 Ω current? A. It is in series with the 1 Ω B. Current sources are open circuits C. It cancels the 12 V source D. It is across L and R, which the batteries hold at 12 V and 6 V Answer: D. Node voltages are fixed by the batteries, so the source only adds to the outer path. (Key point: Current source) 2. The left loop equation is A. 3 I1 + I2 = 12 B. I1 - 3 I2 = 12 C. 3 I1 - I2 = 12 D. 2 I1 + I2 = 12 Answer: C. 12 = 2 I1 + (I1 - I2). (Key point: Left loop) 3. The right loop equation is A. -I1 + 3 I2 = 6 B. -I1 + 3 I2 = -6 C. I1 + 3 I2 = -6 D. 3 I1 - I2 = -6 Answer: B. 0 = 2 I2 + 6 + (I2 - I1), since 6 V is a drop going downward. (Key point: Right loop) 4. The current in the 1 Ω is A. I1 - I2 = 4.5 A downwards B. I1 + I2 = 3 A C. 3.75 A D. 0.75 A Answer: A. 3.75 - (-0.75) = 4.5 A, confirmed by nodal V_M = 4.5 V. 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