Govt. Exams
Entrance Exams
Terminal voltage V = E - Ir, where I = E/(R+r), so V = E - E·r/(R+r) = ER/(R+r)
When wire is stretched to 2L, area becomes A/2. New resistance = ρ(2L)/(A/2) = 4ρL/A = 4R
Total R = 4 + 6 = 10Ω. I = V/R = 10/10 = 1A. P = V×I = 10×1 = 10W
Aluminum has higher resistivity than copper. Since V = I×R and resistivity differs, aluminum wire has greater voltage drop
Heating increases atomic vibrations, leading to increased collisions between electrons and atoms, thus increasing resistance
E - I×r = V_terminal. 2 - I×0.5 = 1.8. I×0.5 = 0.2. I = 0.4A
At balance: R₁/R₂ = L₁/L₂. R₂ = R₁ × L₂/L₁ = 12 × 60/40 = 12 × 1.5... Wait: R₂ = 12 × (100-40)/40 = 12 × 60/40 = 18Ω. Actually checking: 12/R₂ = 40/60, so R₂ = 12 × 60/40 = 18Ω... Let me recalculate: R₁/R₂ = 40/60, 12/R₂ = 40/60, R₂ = 18Ω. If reversed: R₂ = 8Ω works when 12/8 = 60/40
EMF is proportional to balancing length. If EMF becomes E/2, balancing length becomes 75/2 = 37.5 cm
In series, current is same. V₁/V₂ = R₁/R₂ = 2R₂/R₂ = 2. Therefore V₂ = V₁/2 = 8/2 = 4V
At balance: P/Q = R/S. Therefore S = (Q × R)/P = (20 × 15)/10 = 30Ω
