Figure 27-67 displays two circuits with a charged capacitor that is to be discharged through a resistor when a switch is closed. In Fig. 27-67a,${\mathit{R}}_{\mathbf{2}}\mathbf{=}\mathbf{10}\mathit{\Omega }$and ${\mathit{C}}_{\mathbf{2}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{00}\mathit{\mu }\mathit{F}$. In Fig. 27-67b, and. The ratio of the initial charges on the two capacitors is ${\mathit{q}}_{\mathbf{02}}\mathbf{/}{\mathit{q}}_{\mathbf{01}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{50}$. At time t = 0, both switches are closed. At what time t do the two capacitors have the same charge?

• a) The given resistance and capacitance values are:$R_1=20\Omega ,C_1=5\mu F,R_2=10\Omega ,C_2=8\mu F$
• b) The ratio of the initial charges on the two capacitors,$q_{02}/q_{01}=1.50$

Using the concept of charging and discharging a given combination of the resistor and capacitor, we can see that an amount of charge is stored within the given capacitors. Now, using the ratio of the initial charge values, we can get the equation for the condition of the equal charge. By solving it, the required time can be calculated.

Formulae:

The charge equation of a RC circuit, $q=q_0{e}^{-t/\tau }$ (i)

The time constant of the RC circuit, $\tau =RC$ (ii)

The time constant of the capacitor 1 can be given using the given values in equation (ii) as follows:

$$\begin{gathered} {\tau }_1=\left(20\Omega \right)\left(5\times {10}^{-6}\text{F}\right) \\ =1.00\times {10}^{-4}\text{s} \end{gathered}$$

The time constant of the capacitor 2 can be given using the given values in equation (ii) as follows:

$$\begin{gathered} {\tau }_2=\left(10\Omega \right)\left(8\times {10}^{-6}\text{F}\right) \\ =8.00\times {10}^{-5}\text{s} \end{gathered}$$

Now, for the given condition of equal charges, the time value can be calculated using the above values in equation (i) as follows:

$$\begin{gathered} q_1=q_2 \\ {q}_{01}{e}^{-t/{\tau }_1}=q_{02}{e}^{-t/{\tau }_2} \\ {e}^{-t(1/{\tau }_1-1/{\tau }_2)}=q_{02}/q_{01} \\ {e}^{t\left(\frac{{\tau }_1-{\tau }_2}{{\tau }_1{\tau }_2}\right)}=q_{02}/q_{01} \end{gathered}$$

$$\begin{gathered} t=\left(\frac{{\tau }_1{\tau }_2}{{\tau }_1-{\tau }_2}\right)\ln \left(\frac{q_{02}}{q_{01}}\right) \\ =\left(\frac{\left(1.00\times {10}^{-4}\text{s}\right)\left(8.00\times {10}^{-5}\text{s}\right)}{1.00\times {10}^{-4}\text{s}-8.00\times {10}^{-5}\text{s}}\right)\ln \left(\frac{3}{2}\right) \\ =\left(\frac{8\times {10}^{-9}\text{s}}{2\times {10}^{-5}}\right)\left(0.4055\right) \\ =1.62\times {10}^{-4}\text{s} \end{gathered}$$

Hence, the required time is$1.62\times {10}^{-4}\text{s}$