Question 5.15: A circuit with resistance and capacitance is shown in Fig. 5......

At t = 0, the capacitor will be open circuited. In this condition, the Thevenin resistance and voltage can be determined as,

R_{\mathrm{Th}}=1+\left[4+\left({\frac{3\times6}{9}}\right)\right]\,||\,2=1+{\frac{2\times6}{2+6}}=2.5\,\Omega        (5.254)

V_{\mathrm{Th}}=\frac{30}{4+2+2}\times(4+2)=22.5\,\mathrm{V}        (5.255)

The charging voltage is,

\nu_{c}=V_{\mathrm{Th}}\Bigl(1-\mathrm{e}^{-\frac{1}{^{R}Th^{C}}t}\Bigr)=22.5\left(1-e^{-\frac{1}{2.5\times 5}t } \right) =22.5(1-\mathrm{e}^{-0.08t})\,\mathrm{V}        (5.256)