Question

In the circuit shown, the capacitor C₁ and C₂ have capacitance each. The switch S is closed at time t = 0. Taking Q₀ = CE and t = RC, the charge on C₂ after time t will be –

• A. Q₀ (1 –$e^{\frac{- t}{2\tau}}$)
• B. Q₀ (1 –$e^{\frac{- t}{\tau}}$)
• C. $\frac{Q_{0}}{2}(1 - e^{\frac{- t}{2\tau}})$
• D. $\frac{Q_{0}}{2}(1 - e^{\frac{- t}{\tau}})$

Answer 1

Answer: (B)

Q₀ (1 –$e^{\frac{- t}{\tau}}$)

Answer 2

Q = Q₀ (1 –$e^{\frac{- t}{RC_{2}}}$)

= Q₀ (1 –$e^{\frac{- t}{RC}}$)

= Q₀ (1 –$e^{\frac{- t}{\tau}}$)