Govt. Exams
In DC circuits, current is continuous (same) through all series elements including capacitors in steady state. However, in AC circuits, current flows through capacitors.
V = E - Ir. 9 = 10 - I×r. Current I = V/R = 9/90 = 0.1A. So 9 = 10 - 0.1r, giving r = 10Ω
In meter bridge: R₁/R₂ = l₁/(100-l₁). So 8/R₂ = 40/60, giving R₂ = 12Ω
Heat dissipated H = I²Rt. Since same current flows through both and nichrome has higher resistivity than copper (higher R), nichrome dissipates more heat.
E₁/E₂ = l₁/l₂. So E₂ = E₁(l₂/l₁) = 1.5(80/60) = 2V
Terminal voltage V = E - Ir, where I is the current through the circuit. The voltage drop occurs only across internal resistance.
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)
