--- title: "Kirchhoff law currents and source power" subject: "Basic Electrical Engineering" chapter: 2 key: "semester_1/basic_electrical_engineering:kirchhoff-i1-i2-i3-power-output-sources" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/kirchhoff-i1-i2-i3-power-output-sources site: BCT Engineering --- # Kirchhoff law currents and source power Basic Electrical Engineering · Chapter 2: DC circuits - Asked in: 2069 Chaitra, 8 marks - Video: [Using kirchhoff's rules find the current in each resistor shown in figure](https://www.youtube.com/watch?v=tBb5aJaudMM) — WNY 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/kirchhoff-i1-i2-i3-power-output-sources - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2 ## Question Find the currents I1, I2, I3 using Kirchhoff's Law and also find the power output of each voltage source of figure below? [Figure: 10 V source with 2 Ω on the left; 3 Ω on top carrying I1; 7 Ω in the upper middle carrying I2; 3 Ω and a 5 V source on the right; 5 Ω resistors in the shunt and lower branches with I3; another 5 V source at the bottom right.] ## Short answer I₁ = 0.542 A, I₂ = 0.167 A, I₃ = 0.042 A. Power outputs: 10 V source 5.42 W, upper 5 V source 0.83 W, lower 5 V source 1.04 W. ## Answer #### Reading the figure Node P (top middle) joins the 3 Ω (from the left branch), the 7 Ω and the vertical 5 Ω. Node M is the bottom of that 5 Ω; from M, one 5 Ω goes to the ground rail. Node S is the right-hand node, joined to M by a 5 Ω. - Left branch: 10 V battery (+ on top) with 2 Ω and 3 Ω into P. Current I₁ flows into P. - Right top: 7 Ω from node Q to P. Q connects through 3 Ω and a 5 V battery (+ on top) to S. Current I₂ flows from Q to P. - S to M: 5 Ω carrying I₃. - S to ground: 5 Ω with a 5 V battery (+ on top). Polarities read from the plate lengths in the scan. #### Currents by KCL - 5 Ω from P to M: I₁ + I₂ - 5 Ω from M to ground: I₁ + I₂ + I₃ - Lower right branch (towards S): I₂ + I₃ #### KVL equations Left loop (10 V, 2 + 3 Ω, 5 Ω, 5 Ω): 10 = 5 I₁ + 5 (I₁ + I₂) + 5 (I₁ + I₂ + I₃) ⇒ 3 I₁ + 2 I₂ + I₃ = 2 ... (1) Middle loop (P, Q, 3 Ω, 5 V, S, M): 7 I₂ + 3 I₂ − 5 − 5 I₃ + 5 (I₁ + I₂) = 0 ⇒ I₁ + 3 I₂ − I₃ = 1 ... (2) Lower loop (5 V, 5 Ω, S–M 5 Ω, 5 Ω to ground): 5 = 5 (I₂ + I₃) + 5 I₃ + 5 (I₁ + I₂ + I₃) ⇒ I₁ + 2 I₂ + 3 I₃ = 1 ... (3) #### Solution From (2) and (3): subtracting gives I₂ = 4 I₃. Then (1) and (2) give I₃ = 1/24. **I₁ = 13/24 = 0.542 A, I₂ = 1/6 = 0.167 A, I₃ = 1/24 = 0.042 A** #### Power output of each source | Source | Current out of + terminal | Power delivered | |---|---|---| | 10 V | I₁ = 0.542 A | 10 × 0.542 = 5.42 W | | 5 V (upper) | I₂ = 0.167 A | 5 × 0.167 = 0.83 W | | 5 V (lower) | I₂ + I₃ = 0.208 A | 5 × 0.208 = 1.04 W | All three sources deliver power; total = 7.29 W. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The 5 Ω from P to M carries A. I1 - I2 B. I3 C. I1 D. I1 + I2 Answer: D. I1 and I2 both flow into P. (Key point: KCL) 2. Subtracting (3) from (2) gives A. I2 = I3 B. I1 = 4 I3 C. I2 = 4 I3 D. I2 = I3/4 Answer: C. (2) - (3): I2 - 4 I3 = 0. (Key point: Elimination) 3. The solution is A. I1 = 1/24, I2 = 1/6, I3 = 13/24 B. I1 = 13/24 A, I2 = 1/6 A, I3 = 1/24 A C. I1 = 1 A, I2 = 0.5 A, I3 = 0.1 A D. I1 = 13/24, I2 = 1/24, I3 = 1/6 Answer: B. These satisfy all three equations. (Key point: Currents) 4. The lower 5 V source delivers A. 5 x (I2 + I3) = 1.04 W B. 5 x I3 = 0.21 W C. 5 x I1 = 2.7 W D. 5 x I2 = 0.83 W Answer: A. Power = V x current leaving the + terminal. 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