Govt. Exams
Low resistance ammeter minimizes voltage drop in series. High resistance voltmeter draws negligible current in parallel, not affecting the original circuit
In series, same current flows through both. Power = I²R. The 60W bulb has higher resistance (rated at lower power), so it dissipates more power and glows brighter
R = ρL/A. New R = ρ(2L)/π(2r)² = ρ(2L)/(4πr²) = R/2
Each part has resistance R/n. In parallel: 1/R_eq = n/(R/n) = n²/R, so R_eq = R/n²
Current depends on both the resistor value and the rest of the circuit configuration (series/parallel). Without knowing the circuit configuration, it cannot be determined
By potentiometer principle: E₁/E₂ = l₁/l₂, so E₂ = 1.5 × (65/40) = 2.4375V ≈ 2.4V
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
