What is the time constant for the discharge of the capacitors in FIGURE EX28.34?

We have to find that is the time constant for the discharge of the capacitors ?

The switch is off, the capacitors discharge and the current flows through the two resistors. The time taken to discharge the capacitor is called the time constant $\tau$and it is given by equation in the form

$\tau =RC$

$R$is the resistance, and $C$is the capacitance in the circuit. We have two resistors in parallel. If we have two points and there are resistors connected at both ends of the points, we called that the resistors are connected in parallel. This is obvious when the resistors are aligned side by side. For parallel resistors, the equivalent resistance $R_{eq}$is calculated by equation in the form

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+....+\frac{1}{R_N}\left(2\right)$

We use the values for $R_2=R_1=1k\Omega$into equation $\left(2\right)$to get the equivalent resistance for the resistors combination

$$\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2} \\ \frac{1}{R_{eq}}=\frac{1}{1000\Omega }+\frac{1}{1000\Omega } \\ {R}_{eq}={\left(\frac{1}{1000\Omega }+\frac{1}{1000\Omega }\right)}^{-1} \\ {R}_{eq}=500\Omega \end{gathered}$$

The two capacitors are connected in parallels. If we have two points and there are capacitors connected at both ends of the points, we called that the capacitors are connected in parallel. This is obvious when the capacitors are aligned side by side. For parallel capacitors, the equivalent capacitance $C_{eq}$is calculated by

$$\begin{gathered} \frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2} \\ \frac{1}{C_{eq}}=\frac{1}{2\mu F}+\frac{1}{2\mu F} \\ {C}_{eq}={\left(\frac{1}{2\mu F}+\frac{1}{2\mu F}\right)}^{-1} \\ {C}_{eq}=1\mu F \end{gathered}$$

putting the values for $C_{eq}$and $R_{eq}$into equation $\left(1\right)$to get the time constant $\tau$

$$\begin{gathered} \tau =R^{eq}{C}^{eq} \\ =\left(500\Omega \right)\left(1\times {10}^{-6}F\right) \\ =0.5\times {10}^{-3}s \\ =0.5ms \end{gathered}$$