GATE Electrical Engineering (EE) 2021 Solved Papers

During ${\mathit{T}}_{\mathit{o}\mathit{n}}$ :
${\mathit{V}}_{\mathit{i}}=5{\mathit{V}}^{\prime }\mathit{C}$ ' will charge.
${\mathit{V}}_{\mathit{o}}={\mathit{V}}_{\mathit{i}}-{\mathit{V}}_{\mathit{c}}=5-{\mathit{V}}_{\mathit{c}}$
when ${\mathit{V}}_{\mathit{c}}$ increases ${\mathit{V}}_{\mathit{R}}={\mathit{V}}_{\mathit{o}}$ decreases exponentially.
${\mathit{V}}_{\mathit{c}}=\mathit{V}\left(\mathit{\infty }\right)-\left[\mathit{V}\left(\mathit{\infty }\right)-\mathit{V}\left(0\right)\right]{\mathit{e}}^{-\mathit{t}/\mathit{\tau }}$.....(1)
${\mathit{V}}_{\mathit{c}}=5-\left(5-{\mathit{V}}_{\mathit{m}\mathit{i}\mathit{n}}\right){\mathit{e}}^{-\mathit{t}/\mathit{\tau }}$.....(2)
at
${\mathit{V}}_{\max }=5-\left[5-{\mathit{V}}_{\min }\right]{\mathit{e}}^{-\mathit{T}/2\mathit{\tau }}$....(3)
Solving of o/p
${\mathit{V}}_{\mathit{p}-\mathit{p}}=5-{\mathit{V}}_{\min }-\left(-{\mathit{V}}_{\max }\right)=6.2$
$\left({\mathit{V}}_{\max }-{\mathit{V}}_{\min }\right)=1.2\mathit{V}$.......(4)
During ' will discharge towards zero from eq(1)
${\mathit{V}}_{\mathit{c}}=0-\left[0-{\mathit{V}}_{\max }\right]{\mathit{e}}^{-\mathit{t}/\mathit{\tau }}$
${\mathit{V}}_{\mathit{c}}={\mathit{V}}_{\max }{\mathit{e}}^{-\mathit{t}/\mathit{\tau }}$.....(5)
$\mathit{A}\mathit{t}\mathit{t}={\mathit{T}}_{\mathit{o}\mathit{f}\mathit{f}}=\mathit{T}/2{\mathit{V}}_{\mathit{c}}={\mathit{V}}_{\mathit{m}\mathit{i}\mathit{n}}$
${\mathit{V}}_{\min }={\mathit{V}}_{\max }{\mathit{e}}^{-\mathit{T}/2\mathit{\tau }}$.....(6)
We have to find $\mathit{\tau }=\mathit{R}\mathit{C}\Rightarrow \mathit{C}=\mathit{\tau }/\mathit{R}.$..........(7)
Putting the value of eq. 5 in eq. 3 ,
${\mathit{V}}_{\max }=5-\left[5-{\mathit{V}}_{\max }{\mathit{e}}^{-\mathit{T}/2\mathit{\tau }}\right]{\mathit{e}}^{-\mathit{T}/2\mathit{\tau }}$
${\mathit{V}}_{\max }\left[1-{\mathit{e}}^{-\mathit{T}/\mathit{\tau }}\right]=5\left[1-{\mathit{e}}^{-\mathit{T}/2\mathit{\tau }}\right]$
$\left[\because {\mathit{a}}^2-{\mathit{b}}^2=\left(\mathit{a}+\mathit{b}\right)\left(\mathit{a}-\mathit{b}\right)\right]$
${\mathit{V}}_{\max }=\frac{5}{\left[1+{\mathit{e}}^{-\mathit{T}/2\mathit{\tau }}\right]}$
Putting the value of eq. 6 in eq. 4 ,
${\mathit{V}}_{\max }-{\mathit{V}}_{\max }{\mathit{e}}^{-\mathit{T}/2\mathit{\tau }}=1.2$
${\mathit{V}}_{\max }\left[1-{\mathit{e}}^{-\mathit{T}/2\mathit{\tau }}\right]=1.2$
Putting eq. 8 in above equation,
$\left[\frac{5\left[1-{\mathit{e}}^{-\mathit{T}/2\mathit{\tau }}\right]}{\left[1+{\mathit{e}}^{-\mathit{T}/2\mathit{\tau }}\right]}\right]=1.2$
$5-5{\mathit{e}}^{-\mathit{T}/2\mathit{\tau }}=1.2+1.2{\mathit{e}}^{-\mathit{T}/2\mathit{\tau }}$
$3.8=6.2{\mathit{e}}^{-\mathit{T}/2\mathit{\tau }}$
${\mathit{e}}^{\mathit{T}/2\mathit{\tau }}=\frac{6.2}{3.8}$
$\frac{\mathit{T}}{2\mathit{\tau }}=\ln \left(\frac{6.2}{3.8}\right)=0.489$
$\mathit{\tau }=\frac{\mathit{T}}{2\times 0.489}=0.0102\mathit{s}\mathit{e}\mathit{c}$
$\mathit{\tau }=\mathit{R}\mathit{C}$
$\mathit{C}=\frac{\mathit{\tau }}{\mathit{R}}=\frac{0.0102}{820}$
$\mathit{C}=1.245\times {10}^{-5}=12.45\mathit{\mu }\mathit{F}$