Chapter 29, Problem 74PQ

(a)

To determine

The expression for the time constant for this circuit.

Answer to Problem 74PQ

The expression for the time constant is $\underset{\_}{\left(\frac{R_1{R}_2}{R_1+R_2}\right)C}$.

Explanation of Solution

Write the expression for the equivalent resistance for this circuit as.

    $R_{eq}=\frac{R_1{R}_2}{R_1+R_2}$                                                                                                (I)

Here, $R_{eq}$ is the equivalent resistance of circuit, $R_1$ is the resistance of first element and $R_2$ is the resistance of the second element...

Write the expression for the time constant for the circuit parameter.

    $\tau =R_{eq}C$                                                                                                      (II)

Here, $\tau$ is the time constant and $C$ is the capacitor of the circuit.

Conclusion:

Substitute $\frac{R_1{R}_2}{R_1+R_2}$ for $R_{eq}$ in equation (II).

  $\tau =\left(\frac{R_1{R}_2}{R_1+R_2}\right)C$

Thus, the expression for the time constant is $\underset{\_}{\left(\frac{R_1{R}_2}{R_1+R_2}\right)C}$.

(b)

To determine

The expression for the time when the capacitor has lost half of its charge.

Answer to Problem 74PQ

The expression for the time when the capacitor has lost half of its charge is $\underset{\_}{\left(\frac{R_1{R}_2}{R_1+R_2}\left(C\right)\right)\ln \left(2\right)}$.

Explanation of Solution

Write the expression for the charge store in capacitor as.

    $V_c\left(t\right)=\frac{Q\left(t\right)}{C}$                                                                                               (III)

Here, $V_c$ is the voltage across the capacitor, $Q$ is the charge store in capacitor and $C$ is the capacitance of the capacitor

Write the expression for the initial charge stored in capacitor as.

    $V_0=\frac{Q_0}{C}$                                                                                                      (IV)

Here, $Q_o$ is the initial charge stored in the circuit.

Write the expression for voltage across capacitor as a function of time as.

    $V_c\left(t\right)=V_0{e}^{-t/\tau }$                                                                                               (V)

Here, $V_0$ is the initial voltage on the capacitor, $V_c(t)$ is the voltage after time $t$, $R$ is the resistance, C is capacitance and $t$ is time.

Conclusion:

Substitute $\frac{Q\left(t\right)}{C}$ for $V_c\left(t\right)$ and $\frac{Q_0}{C}$ for $V_0$ in equation (V).

    $\begin{array}{l}\frac{Q\left(t\right)}{C}=\frac{Q_0}{C}{e}^{-t/\tau }\\ Q\left(t\right)=Q_0{e}^{-t/\tau }\end{array}$

Substitute $\frac{Q_0}{2}$ for $Q\left(t\right)$ and $R_{eq}C$ for $\tau$ in the above equation.

  $\begin{array}{l}\frac{Q_0}{2}=Q_0{e}^{-t/R_{eq}C}\\ \frac{1}{2}=e^{-t/R_{eq}C}\\ \ln \left(\frac{1}{2}\right)=\frac{-t}{R_{eq}C}\end{array}$

Substitute $\frac{R_1{R}_2}{R_1+R_2}$ for $R_{eq}$ in above equation.

  $t=\left(\frac{R_1{R}_2}{R_1+R_2}\left(C\right)\right)\ln \left(2\right)$

Thus, the expression for the time when the capacitor has lost half of its charge is $\underset{\_}{\left(\frac{R_1{R}_2}{R_1+R_2}\left(C\right)\right)\ln \left(2\right)}$.

(c)

To determine

The expression for the current through the capacitor at that time.

Answer to Problem 74PQ

The expression for the current through the capacitor at that time is $\underset{\_}{-\frac{\left(R_1+R_2\right)V_0}{\left(R_1{R}_2\right)}\left(\frac{1}{2}\right)}$.

Explanation of Solution

Write the expression current through the capacitor at time $t$ as.

    $I\left(t\right)=-\frac{V_0}{R_{eq}}{e}^{-t/\tau }$                                                                                         (VI)

Conclusion:

Substitute $\left(\frac{R_1{R}_2}{R_1+R_2}\left(C\right)\right)\ln \left(2\right)$ for $t$, and $R_{eq}C$ for $\tau$,

  $\begin{array}{c}I\left(t\right)=-\frac{V_0}{R_{eq}}{e}^{-\left(\left(\frac{R_1{R}_2}{R_1+R_2}\left(C\right)\right)\ln \left(2\right)\right)/\left(R_{eq}C\right)}\\ =-\frac{V_0}{R_{eq}}{e}^{-\left(\left(\frac{R_1{R}_2}{R_1+R_2}\right)\ln \left(2\right)\right)/\left(R_{eq}\right)}\end{array}$

Substitute $\frac{R_1{R}_2}{R_1+R_2}$ for $R_{eq}$.in above equation.

  $\begin{array}{c}I(t)=-\frac{V_0}{\left(\frac{R_1{R}_2}{R_1+R_2}\right)}{e}^{-\left(\left(\frac{R_1{R}_2}{R_1+R_2}\right)\ln \left(2\right)\right)/\left(\frac{R_1{R}_2}{R_1+R_2}\right)}\\ =-\frac{V_0}{\left(\frac{R_1{R}_2}{R_1+R_2}\right)}{e}^{-\ln \left(2\right)}\\ =-\frac{\left(R_1+R_2\right)V_0}{\left(R_1{R}_2\right)}\left(\frac{1}{2}\right)\end{array}$

Thus, the expression for the current through the capacitor at that time is $\underset{\_}{-\frac{\left(R_1+R_2\right)V_0}{\left(R_1{R}_2\right)}\left(\frac{1}{2}\right)}$.