--- title: "Mesh current method with current source" subject: "Basic Electrical Engineering" chapter: 2 key: "semester_1/basic_electrical_engineering:mesh-current-10-2a-3-50v" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/mesh-current-10-2a-3-50v site: BCT Engineering --- # Mesh current method with current source Basic Electrical Engineering · Chapter 2: DC circuits - Asked in: 2074 Asoj, 6 marks - Video: [Mesh Analysis with Current Source](https://www.youtube.com/watch?v=s5svOTeKnOE) — Neso Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/mesh-current-10-2a-3-50v - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2 ## Question Find all branch currents in the given circuit by using mesh current method. [Figure: 50 V source on the left; 10 Ω and 5 Ω resistors in the left vertical branch; 2 A current source in series with a 3 Ω resistor in the middle horizontal branch; 2 Ω and 1 Ω resistors in the right vertical branch.] ## Short answer With meshes I1 (left), I2 (upper right), I3 (lower right), constraint I2−I3 = 2 A. Supermesh equations yield I1=20 A, I2=17.33 A, I3=15.33 A. Branches: battery 20 A, 10Ω 2.67 A, 5Ω 4.67 A, 2Ω 17.33 A, 1Ω 15.33 A, source branch 2 A. ## Answer #### Reading the figure Battery 50 V (+ at top) feeds the top rail T. Branches: 10 Ω from T to the middle node M, 5 Ω from M to the bottom rail, 2 Ω from T to node N, 1 Ω from N to the bottom rail. A 2 A source (with the 3 Ω in series) joins N to M, direction N to M. The 3 Ω carries 2 A whatever else happens, so it does not appear in the equations. #### Mesh currents (all clockwise) - I₁: left mesh (50 V, 10 Ω, 5 Ω) - I₂: upper right mesh (10 Ω, 2 Ω, source branch) - I₃: lower right mesh (5 Ω, source branch, 1 Ω) #### Equations Current source branch (flows N to M, i.e. I₂ minus I₃): I₂ − I₃ = 2 Mesh 1: 50 = 10 (I₁ − I₂) + 5 (I₁ − I₃) ⇒ 15 I₁ − 10 I₂ − 5 I₃ = 50 Supermesh (meshes 2 and 3 combined, because the source voltage is unknown): 2 I₂ + 1 I₃ + 10 (I₂ − I₁) + 5 (I₃ − I₁) = 0 ⇒ −15 I₁ + 12 I₂ + 6 I₃ = 0 #### Solution Put I₃ = I₂ − 2: 15 I₁ − 15 I₂ = 40 and −15 I₁ + 18 I₂ = 12 Adding: 3 I₂ = 52, so I₂ = 17.33 A; I₁ = 20 A; I₃ = 15.33 A #### Branch currents - Battery: 20 A - 10 Ω: I₁ − I₂ = 2.67 A (T to M) - 5 Ω: I₁ − I₃ = 4.67 A (M to ground) - 2 Ω: 17.33 A (T to N) - 1 Ω: 15.33 A (N to ground) - 3 Ω and current source: 2 A (N to M) **Answer: 20 A, 2.67 A, 4.67 A, 17.33 A, 15.33 A and 2 A as above.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the 2 A source branch treated in the mesh equations? A. As a 2 ohm resistor B. It fixes the difference of the two mesh currents sharing it: I2 - I3 = 2 C. It is ignored completely D. It sets I1 = 2 A Answer: B. The branch current is fixed by the source, giving a constraint equation between meshes 2 and 3. (Key point: Current source branch) 2. Why is a supermesh used for meshes 2 and 3? A. The voltage across the current source is unknown, so the two meshes are combined with KVL around the outside B. They have the same resistors C. The battery lies in both meshes D. To avoid using KVL Answer: A. An unknown source voltage cannot be written in a mesh equation, so the loop is taken around both meshes avoiding the source branch. (Key point: Supermesh) 3. Why does the 3 ohm resistor not appear in the equations? A. It is zero ohm B. It is shorted C. It carries 2 A fixed by the series current source, and the supermesh avoids that branch D. It carries no current Answer: C. Its current is fixed at 2 A by the source, so it only affects the unknown source voltage. (Key point: 3 ohm resistor) 4. The battery current (I1) and the 10 ohm current are: A. 20 A and 17.33 A B. 15.33 A and 2.67 A C. 2 A and 4.67 A D. 20 A and 2.67 A Answer: D. I1 = 20 A and the 10 ohm carries I1 - I2 = 20 - 17.33 = 2.67 A. 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