Practice question

Given: In a Wheatstone bridge with R_1 = 30 Ω, R_2 = 60 Ω, R_3 = 15 Ω, and R_4 = 32 Ω, a 10 V battery is connected across AC. What is the current through the galvanometer ( R_g = 10 Ω )? Formula: Junction B: I_1 = I_g + I_4, Junction D: I_2 = I_g + I_3. Substitution & Calculation: Apply Kirchhoff’s rules. Let currents be I_1 (AB), I_2 (AD), I_g (BD). . Loop BADB: 30 I_1 + 10 I_g - 60 I_2 = 0 Rightarrow 3 I_1 + I_g - 6 I_2 = 0 . Loop BCDB: 60 (I_1 - I_g) - 10 I_g - 15 (I_2 + I_g) = 0 Rightarrow 4 I_1 - 2 I_2 - 2 I_g = 0 . Loop ADCEA: 60 I_2 + 15 (I_2 + I_g) = 10 Rightarrow 75 I_2 + 15 I_g = 10 Rightarrow 5 I_2 + I_g = 2/3 . Solve: From (3) I_g = 2/3 - 5 I_2, substitute in (1): 3 I_1 + 2/3 - 5 I_2 - 6 I_2 = 0 Rightarrow 3 I_1 - 11 I_2 = -2/3 . From (2): 4 I_1 - 2 I_2 - 2 (2/3 - 5 I_2) = 0 Rightarrow 4 I_1 - 2 I_2 - 4/3 + 10 I_2 = 0 Rightarrow 4 I_1 + 8 I_2 = 4/3 . Solve: 12 I_1 - 33 I_2 = -2, 12 I_1 + 24 I_2 = 4 . Subtract: -57 I_2 = -6 Rightarrow I_2 = 6/57 = 2/19 A . I_g = 2/3 - 5 × 2/19 = 38/57 - 30/57 = 8/57 approx 0.14 A . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.