Govt. Exams
Entrance Exams
The bridge is balanced when P/Q = R/S. In balanced condition, potential difference across galvanometer is zero, hence no current flows through it.
When stretched to 2L, volume remains constant. New area A' = A/2. New resistance R' = ρ(2L)/(A/2) = 4ρL/A = 4R
Parallel: 1/R_p = 1/100 + 1/200 = 3/200, so R_p = 66.67Ω. Total = 50 + 66.67 = 116.67Ω
Taking one direction as positive: 2 + 4 - 3 = 3V or 2 - 4 + 3 = 1V. Net EMF = 1V (assuming standard configuration)
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
