A capacitor of capacity 0.1 F connected in series to a resistor of 10 M is charged to a certain potential and then made to discharge through the resistor. The time in which the potential will take to fall to half its original value is (Given, ₁₀ 2=0.3010 )

Step-by-step solution

By equation of charging q=q₀ (1-e^ -t / C R ) According to question q q₀ = 1 2 =0.50 0.50=1-e^ -t / C R e^ -t / C R =1-0.50=0.50e^ t / C R =2 or t C R = _e 2 or t C R =2.3026 ₁₀ 2 or t=C R 2.3026 ₁₀ 2 or t=0.1 10⁻⁶ 10 10^6 2.3026 ₁₀ 2 or t=2.3026 0.3010 or t=0.693 ~s