Govt. Exams
Entrance 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
