( 6 = 1I_2 + 3(I_2 - I_1) ) ( 6 = 4I_2 - 3I_1 ) … (2)
1. Fundamental Principles Q1: State Kirchhoff’s First Law (Current Law). A1: The sum of currents entering any junction is equal to the sum of currents leaving that junction. [ \sum I_\textin = \sum I_\textout ] Explanation: Based on conservation of charge. kirchhoff 39-s laws questions and answers pdf a level
Solve (1) and (2): From (2): ( 3I_1 = 4I_2 - 6 ) → ( I_1 = \frac4I_2 - 63 ) Sub into (1): ( 12 = 5 \cdot \frac4I_2 - 63 - 3I_2 ) Multiply by 3: ( 36 = 20I_2 - 30 - 9I_2 ) ( 66 = 11I_2 ) → ( I_2 = 6 , \textA ) Then ( I_1 = \frac4\times 6 - 63 = \frac183 = 6 , \textA ) ( 6 = 1I_2 + 3(I_2 - I_1) ) ( 6 = 4I_2 - 3I_1 ) … (2) 1
(a) Combine A+R₁ branch: 12V + 1Ω + 2Ω = 12V, 3Ω total. B+R₂ branch: 8V + 1Ω + 4Ω = 8V, 5Ω total. These two branches in parallel with R₃=2Ω. Use superposition or loop equations for accurate answer. Final equivalent resistance ≈ 1.67Ω. [ \sum I_\textin = \sum I_\textout ] Explanation: