Let initial balance condition be: R₁ + 10/R₂ = 50/50 After removing 10 Ω, new balance condition: R₁/R₂ = 40/60 Solving for R₁, we get 40 Ω. Answer: (c) 40 Ω.
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For series connection: Total resistance Rₛ = nR + R, current I = E / (nR + R). For parallel connection: Equivalent resistance Rₚ = R/n, total resistance Rₜ = (R/n) + R, current 10I = E / [(R/n) + ...
Resistance R of a wire is R = ρ (l / A) = ρ (l / πr²). Given l₁:l₂:l₃ = 2:3:4 and r₁:r₂:r₃ = 3:4:5, solving for I, the ratio is 54:64:75. Answer: (b) 54:64:75.
For n identical resistors of resistance R: Series combination: S = nR Parallel combination: P = R/n Given 5 = nP, substituting P = R/n: 5 = n (R/n) ⇒ 5 = R, so n = 4. Answer: (d) 4.
Total resistance Rₜ = 21Ω + 4Ω = 25Ω. Using V = IR = 0.2 × 25 = 5V. Power P = VI = 5 × 0.2 = 1W = 1 J/s. Answer: (d) 1 J/s.