In Fig. 8.23, both switches are open for t

Question

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.

Accepted Answer

We use (8.23) (twice). We are given [latex]ν_C\left(0^{-} ight)=0 [/latex], which implies [latex]ν_C\left(0^{+} ight)=0 [/latex], because the voltage across a capacitor is continuous. For [latex]0

[latex] ν(t)=ν(\infty)-\left[ν(\infty)-ν\left(0^{+} ight) ight] \exp (-t / au), t>0 [/latex] (8.23)

[latex] au_1=\left(R_1+R_2 ight) C . [/latex]

Thus for [latex] 0

[latex] \begin{aligned} ν_C(t) &=V_0\left[1-\exp \left(-\frac{t}{ au_1} ight) ight]\\ au_1 &=\left(R_1+R_2 ight) C, 0

For [latex] t_1

[latex] au_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^{+} ight)=V_0\left[1-\exp \left(-\frac{t_1}{ au_1} ight) ight] [/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}{ au_1} ight) ight] ight\} \\& \exp \left(-\frac{t-t_1}{ au_2} ight)=V_0-V_0 \ & \exp \left(-\frac{t_1}{ au_1} ight) \exp \left(-\frac{t-t_1}{ au_2} ight) \\ au_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 / au_1 ight) ight], \quad 0t_1 \end{array} ight. [/latex]

where

[latex] au_1=\left(R_1+R_2 ight) C, au_2=R_1 C [/latex]

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

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