Question

Two capacitors of capacitance 2C and C are charged to potential difference 2V and V respectively. Now the charged capacitors are connected with a resistor R as shown in figure. Switch ‘S’ is closed at t = 0.

What is the current in circuit as function of time?

• A. $\frac{V}{R}{e}^{\frac{-3t}{RC}}$
• B. $\frac{2V}{R}{e}^{\frac{-2t}{3RC}}$
• C. $\frac{V}{R}{e}^{\frac{-t}{RC}}$
• D. $\frac{V}{R}{e}^{\frac{-3t}{2RC}}$

Answer 1

The correct option is D

$\frac{V}{R}{e}^{\frac{-3t}{2RC}}$

Let x be the charge flowing in the circuit. $\left(\frac{4CV-x}{2C}\right)-\left(\frac{CV+x}{C}\right)=R.\frac{dx}{dt}$

$\int \frac{dx}{2CV-3x}=\int \frac{dt}{2RC}orx=\frac{2CV}{3}[1-e^{\frac{-3}{2CR}}]ori=\frac{dx}{dt}=\frac{V}{R}{e}^{-\frac{3t}{2CR}}$