Menu

A2: The sum of electromotive forces (e.m.f.) around any closed loop in a circuit equals the sum of potential differences (p.d.) around the same loop. [ \sum \mathcalE = \sum IR ] Explanation: Based on conservation of energy. 2. Simple Circuit Analysis Q3: In a series circuit with a 12V battery, R₁ = 2Ω and R₂ = 4Ω. Find the current and the p.d. across each resistor. A3: Total resistance ( R_T = 2 + 4 = 6 \Omega ) Current ( I = \fracVR_T = \frac126 = 2 , \textA ) ( V_1 = I \times R_1 = 2 \times 2 = 4 , \textV ) ( V_2 = I \times R_2 = 2 \times 4 = 8 , \textV ) Check: ( 4 + 8 = 12 , \textV ) (K2 satisfied)

( 12 = 2I_1 + 3(I_1 - I_2) ) ( 12 = 5I_1 - 3I_2 ) … (1)

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.

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 )

(c) A V Chiptroniks Technology Pvt. Ltd. Coaching WordPress Theme