At t=τ, q=CV/2
At t=2τ, q=CV(1-e⁻²)
At t=τ/2, q=CV(1-e⁻½)
Work done by the battery is half of the energy dissipated in the resistor
At switch S1 is closed and switch S2 is kept open. Now capacitor is charging through a resistor R. Charge on a capacitor at any time t is q=q₀(1-e⁻ᵗ⁄τ) = CV(1-e⁻ᵗ⁄τ) [As q₀=CV] At t=τ/2, q=CV(1-e⁻τ/(2τ))=CV(1-e⁻½) At t=τ, q=CV(1-e⁻τ/τ)=CV(1-e⁻¹) At t=2τ, q=CV(1-e⁻²τ/τ)=CV(1-e⁻²).