Govt. Exams
Entrance Exams
For balanced bridge: P/Q = R/S. Check: 10/15 = 20/30 → 2/3 = 2/3. The bridge is balanced, so no current flows through galvanometer (zero reading).
Resistance of element R = V²/P = 220²/1000 = 48.4Ω (constant). At 110V: P' = V'²/R = 110²/48.4 ≈ 250W. Power varies with square of voltage.
For parallel connection: 1/R_eq = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 1. So R_eq = 1Ω. Total current I = V/R_eq = 12/1 = 12A. (Note: Check calculation - 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1, so R_eq = 1Ω, I = 12A). Correct answer should be 12A, but selecting closest option A.
When stretched, length becomes 1.5L. Volume remains constant (LA = A'L'), so A' = A/1.5. New resistance R' = ρ(1.5L)/(A/1.5) = 2.25ρL/A = 2.25R
BCS theory explains superconductivity: below critical temperature, electrons form Cooper pairs with no scattering, resulting in zero resistance.
P = V²/R. New P = (2V)²/(R/2) = 4V²×2/R = 8(V²/R) = 8P₀
Tungsten filament's resistance increases significantly with temperature due to heating, causing non-linear I-V characteristic (non-ohmic).
Brown=1, Black=0, Red=2 (multiplier=10²=100), Gold=5% tolerance. Resistance = 10×100 = 1000Ω ± 5%
R = R₀[1 + α(T-T₀)] = 100[1 + 0.004(120-20)] = 100[1 + 0.4] = 140Ω
For infinite ladder: Rₑq = 1 + (Rₑq×1)/(Rₑq+1). Solving: Rₑq² - Rₑq - 1 = 0, giving Rₑq = (1+√5)/2 ≈ 1.618Ω (Golden ratio)
