Question 5.18: The switch is opened at t = 0 in the circuit as shown in Fig......
The switch is opened at t = 0 in the circuit as shown in Fig. 5.49. Calculate the voltage across the capacitor at t>0 condition.
Att=0^{-}, the circuit is shown in Fig. 5.50. The source and branch currents are calculated as,
i={\frac{24}{4+{\frac{3×6}{9}}}}=4{\mathrm{~A}} (5.305)
i_{6\Omega}=4\times{\frac{3}{3+6}}=1.33\ \mathrm{A} (5.306)
The voltage across the capacitor is,
\nu_{c}(0^{-})=6\times1.33=7.98\ \mathrm{V} (5.307)
\nu_{c}(0^{-})=\nu_{c}(0^{+})=7.98\ \mathrm{V} (5.308)
At t>0, the circuit is shown in Fig. 5.51 and the steady-state voltage across the capacitor is,
V_{s s}=0 (5.309)
Thevenin resistance and time constant are,
R_{\mathrm{Th}}={\frac{3\times6}{3+6}}=2\,\Omega (5.310)
R_{\mathrm{Th}}C=5\times2=10\,\mathrm{s} (5.311)
At t > 0, the voltage across the capacitor is,
\nu_{c}(t)=V_{s s}+[\nu_{c}(0^{+})-V_{s s}]\mathrm{e}^{\frac{-t}{^{R}Th^{C} }}=0+[7.98-0]\mathrm{e}^{\frac{-t}{10}}=7.98\mathrm{e}^{-0.1t}\mathrm{V} (5.312)
