--- title: "Mesh analysis, power in 15Ω" subject: "Basic Electrical Engineering" chapter: 2 key: "semester_1/basic_electrical_engineering:mesh-power-dissipation-15-ohm-400v-200v" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/mesh-power-dissipation-15-ohm-400v-200v site: BCT Engineering --- # Mesh analysis, power in 15Ω Basic Electrical Engineering · Chapter 2: DC circuits - Asked in: 2072 Chaitra, 6 marks - Video: [Mesh Analysis for Circuits Explained](https://www.youtube.com/watch?v=eQpc2QRFv7Y) — Engineer4Free - 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-power-dissipation-15-ohm-400v-200v - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2 ## Question Find the power dissipation in 15 Ω resistor shown in figure below using mesh analysis. [Figure: top branches 20 Ω, 15 Ω and 10 Ω in series; shunt branches: 400 V source on the left, 80 Ω in the first vertical, 90 Ω in the second vertical, 200 V source on the right.] ## Short answer Define mesh currents I1 (left), I2 (middle), I3 (right). Equations: 100I1 - 80I2 = 400, -80I1 + 185I2 - 90I3 = 0, -90I2 + 100I3 = -200. Solving gives I2 = 3.5 A. Power dissipation in 15 Ω resistor = 3.5² × 15 = 183.75 W. ## Answer #### Reading the figure Top branch (left to right): 20 Ω, 15 Ω, 10 Ω. Vertical branches: 400 V battery (left, + on top), 80 Ω, 90 Ω, 200 V battery (right, + on top). The bottom is a common wire. #### Mesh currents (clockwise) - I₁: left mesh (400 V, 20 Ω, 80 Ω) - I₂: middle mesh (15 Ω, 90 Ω, 80 Ω) - I₃: right mesh (10 Ω, 200 V, 90 Ω) #### Equations Mesh 1: 400 = 20 I₁ + 80 (I₁ − I₂) ⇒ 100 I₁ − 80 I₂ = 400 Mesh 2: 0 = 15 I₂ + 90 (I₂ − I₃) + 80 (I₂ − I₁) ⇒ −80 I₁ + 185 I₂ − 90 I₃ = 0 Mesh 3 (going down through the 200 V battery from + to −, a drop of 200 V): 0 = 10 I₃ + 200 + 90 (I₃ − I₂) ⇒ −90 I₂ + 100 I₃ = −200 #### Solution I₁ = 6.8 A, I₂ = 3.5 A, I₃ = 1.15 A The current in the 15 Ω is I₂ = 3.5 A. #### Power in the 15 Ω P = I₂² × 15 = 3.5² × 15 = 12.25 × 15 = 183.75 W **Answer: P₁₅Ω = 183.75 W** Check: the 80 Ω carries I₁ − I₂ = 3.3 A and the 90 Ω carries I₂ − I₃ = 2.35 A. The power from the 400 V source is 2720 W; the 200 V source absorbs 230 W; the resistors dissipate the rest. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Mesh 1 equation is A. 100 I1 + 80 I2 = 400 B. 20 I1 - 80 I2 = 400 C. 100 I1 - 80 I2 = 0 D. 100 I1 - 80 I2 = 400 Answer: D. 400 = 20 I1 + 80(I1 - I2) = 100 I1 - 80 I2. (Key point: Mesh 1 equation) 2. In mesh 3 the 200 V appears as + 200 on the right because A. it is a voltage rise B. the battery is ignored C. going down through it from + to - is a voltage drop in the clockwise direction D. the 200 V is in the middle mesh Answer: C. The 200 V opposes the clockwise current, giving 0 = 10 I3 + 200 + 90(I3 - I2). (Key point: Mesh 3 sign) 3. The current in the 15 Ω is I2 because A. it is shared by all three meshes B. the 15 Ω is in the middle mesh only (not shared) C. it is I1 - I2 D. it is I3 Answer: B. A resistor in only one mesh carries just that mesh current. (Key point: Current in 15 Ω) 4. With I2 = 3.5 A, the power in 15 Ω is A. I^2 R = 183.75 W B. 3.5 x 15 = 52.5 W C. 3.5^2/15 = 0.82 W D. 6.8^2 x 15 W Answer: A. P = 3.5^2 x 15 = 12.25 x 15 = 183.75 W. 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