in t = 0 3ia 9V 25mF: VС 4.44Н; Find vc(t) for t>0 +

Given: A circuit which is in steady state before t=0 and at t=0 switch is opened.

Formula and concept: At steady state inductor behaves as short circuit and capacitor behaves as open circuit. 

According to question:

Before t=0.

Applying KVL in loop [abcdefa], this will give

$$\begin{gathered} 3i_A+7\left[I-i_A\right]+0-9=0 \\ \Rightarrow 7I-4i_A=9 \\ \& \\ {i}_A=\frac{9}{2} \\ \Rightarrow i_A=4.5\text{A} \\ \Rightarrow I=\frac{9+4\left(4.5\right)}{7} \\ \Rightarrow I=3.85\text{A} \\ \therefore i_L\left(0^{-}\right)=3.85\text{A} \\ \& \\ {V}_C\left(0^{-}\right)=0\text{V} \end{gathered}$$

At t=0,switch is opened and current through inductor and voltage across capacitor cannot change instantaneously.

$$\begin{gathered} \therefore i_L\left(0^{+}\right)=3.85\text{A} \\ \& \\ {V}_C\left(0^{+}\right)=0\text{V} \end{gathered}$$

The circuit now becomes

Applying KVL in loops [abefa] and [bcdeb], this will give

$$\begin{gathered} 7i_A-3i_A+2i_A+\frac{1}{25\times {10}^{-3}}\int i_{1}dt=0 \\ \Rightarrow 6i_A+\frac{1}{25\times {10}^{-3}}\int i_{1}dt=0 \\ \text{Applying Laplace transform} \\ \therefore 6I_A\left(S\right)+40\frac{I_1\left(S\right)}{S}=0...........\left(1\right) \\ 4.44\frac{d{i}_2}{dt}+\frac{1}{25\times {10}^{-3}}\int i_{1}dt=0 \\ \text{Applying Laplace transform} \\ 4.44\left[S{I}_2\left(S\right)-3.85\right]+40\frac{I_1\left(S\right)}{S}=0 \\ \Rightarrow 4.44S{I}_2\left(S\right)+40\frac{I_1\left(S\right)}{S}=17.125............\left(2\right) \\ \& \\ {i}_1=i_A+i_2 \\ \text{Applying Laplace transform} \\ {I}_1\left(S\right)=I_A\left(S\right)+I_2\left(S\right).................\left(3\right) \\ \text{From (1) \& (3)} \\ 6\left[I_1\left(S\right)-I_2\left(S\right)\right]+40\frac{I_1\left(S\right)}{S}=0 \\ \Rightarrow 6I_1\left(S\right)+40\frac{I_1\left(S\right)}{S}=6I_2\left(S\right) \\ \Rightarrow I_1\left(S\right)\left[6+\frac{40}{S}\right]=6I_2\left(S\right) \\ \Rightarrow I_1\left(S\right)=\frac{6S{I}_2\left(S\right)}{6S+40}.............\left(4\right) \\ \text{From (2) \& (4)} \\ 4.44S{I}_2\left(S\right)+40\frac{1}{S}\times \frac{6S{I}_2\left(S\right)}{6S+40}=17.125 \\ \Rightarrow 4.44S{I}_2\left(S\right)+\frac{240I_2\left(S\right)}{6S+40}=17.125 \\ \Rightarrow I_2\left(S\right)\left[4.44S\left(6S+40\right)+240\right]=17.125\left(6S+40\right) \\ \Rightarrow I_2\left(S\right)\left[26.64S^2+177.6S+240\right]=17.125\left(6S+40\right) \\ \Rightarrow I_2\left(S\right)=\frac{17.125\left(6S+40\right)}{\left[26.64S^2+177.6S+240\right]} \\ \Rightarrow I_2\left(S\right)=\frac{3.8568\left(S+6.66\right)}{\left[S^2+6.66S+9\right]} \\ \Rightarrow I_2\left(S\right)=\frac{3.8568(S+6.66)}{\left[\left(S+1.88\right)(S+4.77)\right]} \\ \mathrm{By} \mathrm{partial} \mathrm{fraction} \\ \therefore I_2\left(S\right)=\frac{6.37}{\left(S+1.88\right)}-\frac{2.51}{(S+4.77)} \\ \text{Applying inverse Laplace transform} \\ \therefore i_2\left(t\right)=\left[6.37e^{-1.88t}-2.51e^{-4.77t}\right]\text{A} \\ \because v_L\left(t\right)=4.44\frac{d{i}_2}{dt} \\ \therefore v_L\left(t\right)=4.44\frac{d\left(6.37e^{-1.88t}-2.51e^{-4.77t}\right)}{dt} \\ \Rightarrow v_L\left(t\right)=4.44\left[6.37\times -1.88e^{-1.88t}-2.51\times -4.77e^{-4.77t}\right] \\ \Rightarrow v_L\left(t\right)=\left[53.146e^{-4.77t}-53.146e^{-1.88t}\right]\text{V} \\ \because v_C\left(t\right)=v_L\left(t\right) \\ \therefore v_C\left(t\right)=\left[53.146e^{-4.77t}-53.146e^{-1.88t}\right]\text{V} \end{gathered}$$