Question

What is the time constant for the discharge of the capacitors in FIGURE EX28.33?

Short answer

The time constant for the discharge of the capacitors is $\tau =8ms$

Step-by-step solution

Given Information

We have to find the time constant for the discharge of the capacitors.

Simplify

The time when 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 the equation.

localid="1648651176873" $\tau =RC...\left(1\right)$

where,$R$the resistance, and $C$is the capacitance in the circuit. When the two resistors are connected one by one and with no junctions between them. they are called to be connected in series. For a series of resistors, the equivalent resistance for their combination is given by the equation.

$$\begin{gathered} R_{eq}=R_1+R_2 \\ =1000\Omega +1000\Omega \\ =2000\Omega \end{gathered}$$

The two capacitors are connected in parallel. 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.

$$\begin{gathered} C_{eq}=C_1+C_2 \\ =2\mu F+2\mu F \\ =4\mu F \end{gathered}$$

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

localid="1648648177288" $$\begin{gathered} \tau =R_{eq}{C}_{eq} \\ =\left(2000\Omega \right)(4\times {10}^{-6}F) \\ =8\times {{10}^{-3}}_s \\ =8ms \end{gathered}$$