In Fig. 8.23, both switches are open for t<0 , and [latex] ν_C\left(0^{-}\right)=0 [/latex] . Switch [latex] S_1 [/latex] is closed at t=0 and switch [latex] S_2 [/latex] is closed at [latex] t=t_1 [/latex] . Obtain an expression for the voltage [latex] ν_{ C }(t) [/latex] across the capacitor.

We use (8.23) (twice). We are given [latex]ν_C\left(0^{-}\right)=0 [/latex], which implies [latex]ν_C\left(0^{+}\right)=0 [/latex], because the voltage across a capacitor is continuous. For [latex]0<t \leq t_1[/latex], the voltage [latex]ν_C[/latex] (t) rises toward [latex]V_0[/latex] with time constant given by [latex] ν(t)=ν(\infty)-\left[ν(\infty)-ν\left(0^{+}\right)\right] \exp (-t / \tau), t>0 [/latex] (8.23) [latex] \tau_1=\left(R_1+R_2\right) C . [/latex] Thus for [latex] 0<t \leq t_1, ν(\infty)=V_0 [/latex], and (8.23) becomes [latex] \begin{aligned} ν_C(t) &=V_0\left[1-\exp \left(-\frac{t}{\tau_1}\right)\right]\\\\ \tau_1 &=\left(R_1+R_2\right) C, 0<t \leq t_1 \end{aligned}[/latex] For [latex] t_1<t<\infty [/latex], the resistor [latex]R_2[/latex] is bypassed by a short circuit, so the time constant is given by [latex]\tau_2=R_1 C . [/latex] If we take [latex] t_1 [/latex] as the time origin for the second segment, the initial and final values of [latex] ν_{ C }(t) [/latex] are given by [latex] ν_C\left(0^{+}\right)=V_0\left[1-\exp \left(-\frac{t_1}{\tau_1}\right)\right] [/latex]; [latex] ν_C(\infty)=V_0[/latex]. We then reset the time origin to zero (delay the response by [latex] t_1[/latex] ) and obtain [latex] \begin{aligned} ν_C(t)=& V_0-\left\{V_0-V_0\left[1-\exp \left( \frac{t_1}{\tau_1}\right)\right]\right\} \\\\& \exp \left(-\frac{t-t_1}{\tau_2}\right)=V_0-V_0 \\ & \exp \left(-\frac{t_1}{\tau_1}\right) \exp \left(-\frac{t-t_1}{\tau_2}\right) \\\\ \tau_2=& R_1 C, t>t_1 \end{aligned} [/latex] Therefore [latex] ν_C(t)=V_0\left\{\begin{array}{l} 0, \quad t<0 \\\\ {\left[1-\exp \left(-t / \tau_1\right)\right], \quad 0<t \leq t_1} \\\\ 1-\exp \left(-t / \tau_1\right) \exp \left[-\left(t-t_1\right) / \tau_2\right], \quad t>t_1 \end{array}\right. [/latex] where [latex]\tau_1=\left(R_1+R_2\right) C, \tau_2=R_1 C [/latex]

Question 8.11: In Fig. 8.23, both switches are open for t<0 , and νC(0−...

Introduction to Circuit Analysis and Design

In Fig. 8.23, both switches are open for t<0 , and $ν_C\left(0^{-}\right)=0$ . Switch $S_1$ is closed at t=0 and switch $S_2$ is closed at $t=t_1$ . Obtain an expression for the voltage $ν_{ C }(t)$ across the capacitor.

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We use (8.23) (twice). We are given $ν_C\left(0^{-}\right)=0$ , which implies $ν_C\left(0^{+}\right)=0$ , because the voltage across a capacitor is continuous. For $0<t \leq t_1$ , the voltage $ν_C$ (t) rises toward $V_0$ with time constant given by

$ν(t)=ν(\infty)-\left[ν(\infty)-ν\left(0^{+}\right)\right] \exp (-t / \tau), t>0$ (8.23)

$\tau_1=\left(R_1+R_2\right) C .$

Thus for $0<t \leq t_1, ν(\infty)=V_0$ , and (8.23) becomes

$\begin{aligned} ν_C(t) &=V_0\left[1-\exp \left(-\frac{t}{\tau_1}\right)\right]\\\\ \tau_1 &=\left(R_1+R_2\right) C, 0<t \leq t_1 \end{aligned}$

For $t_1<t<\infty$ , the resistor $R_2$ is bypassed by a short circuit, so the time constant is given by

$\tau_2=R_1 C .$

If we take $t_1$ as the time origin for the second segment, the initial and final values of $ν_{ C }(t)$ are given by

$ν_C\left(0^{+}\right)=V_0\left[1-\exp \left(-\frac{t_1}{\tau_1}\right)\right]$ ;

$ν_C(\infty)=V_0$ .

We then reset the time origin to zero (delay the response by $t_1$ ) and obtain

$\begin{aligned} ν_C(t)=& V_0-\left\{V_0-V_0\left[1-\exp \left( \frac{t_1}{\tau_1}\right)\right]\right\} \\\\& \exp \left(-\frac{t-t_1}{\tau_2}\right)=V_0-V_0 \\ & \exp \left(-\frac{t_1}{\tau_1}\right) \exp \left(-\frac{t-t_1}{\tau_2}\right) \\\\ \tau_2=& R_1 C, t>t_1 \end{aligned}$

Therefore

$ν_C(t)=V_0\left\{\begin{array}{l} 0, \quad t<0 \\\\ {\left[1-\exp \left(-t / \tau_1\right)\right], \quad 0<t \leq t_1} \\\\ 1-\exp \left(-t / \tau_1\right) \exp \left[-\left(t-t_1\right) / \tau_2\right], \quad t>t_1 \end{array}\right.$

where

$\tau_1=\left(R_1+R_2\right) C, \tau_2=R_1 C$