Question

The switch in the network in Fig. E7.17 opens at t=0. Find i(t) for t>0

Answer 1

Solution

For t<0 

capacitor open circuit induction short circuit 

\( V_c(0^{-})=0V \) and \( I_L=\frac{12}{6}= 2A \) => \( I_L(0^-)=2A \)

For t>0

\( \alpha= \frac{R}{2L}=\frac{3}{2*2}=0.75 \)

\( \omega_0=\frac{1}{ \sqrt {LC}}=\frac{1}{\sqrt {2}}=0.7071 \)

\( \alpha > \omega_0 \) => system is ovoi demped 

roots

\( S_{1,2} \) = \( - \alpha \pm \sqrt {\alpha^2- \omega_0^2}=-0.75\pm0.25=-1,-0.5 \)

response is \( i(t)=A_1e^{-t}+A_2e^{-0.5t} \)       (1)

at t=0

(1): \( i(0)=A_1+A_2 \)

\( A_1+A_2=2 \)    (2)

\( V_c(t)=\frac{1}{c} \int i(t)dt \)

          \( =\int ( A_1e^{-t}+A_2e^{-0.5t} )dt \)

\( V_c(t)=\frac{A_1e^{-t}}{-1}+\frac{A_2e^{-0.5t}}{-0.5} \)

\( V_c(t)=-Ae^{-t}-2A_2e^{-0.5t} \)

t=0 => \( V_c(0)=-A_1-2A_2 \)

\( -A_1-2A_2=0 \)  (3)

solve (2) and (3) we get  \( A_1=4,A_2=-2 \)

(1): \( i(t)=4e^{-t}-2e^{-0.5} \)

Answer

Therefore,

For t<0

\( V_c(0^{-})=0 \) , \( I_L({o^-} ) =2A \)

For t>0

\( i(t)=A_1e^{-t}+A_2e^{-0.5t} \)

For t=0

\( i(t)=4e^{-t}-2e^{-0.5t} \)