Question 5.19: Figure 5.54 shows a circuit where the switch is opened for a......
Figure 5.54 shows a circuit where the switch is opened for a long time. Determine the current i(t) for t>0.
At \textstyle t=0^{-}, the switch is still open circuited and the inductor is short circuited. This circuit in this condition is shown in Fig. 5.55 and in this case, the current is,
i_{L}(0^{-})=i_{L}(0^{+})={\frac{16}{3+4}}=2.29\,\mathrm{A} (5.322)
At t=0^{+}, i. e. t>0, the switch is closed and the circuit is shown in Fig. 5.56. The total circuit resistance is,
i=\frac{16}{4+3\mid\mid\left(1+5\right)}=2.67\,\mathrm{A} (5.323)
i_{ss,L}=2.67\times{\frac{(1+5)}{6+3}}=1.78\,\mathrm{A} (5.324)
The Thevenin resistance can be calculated from the circuit as shown in Fig. 5.56 as,
R_{\mathrm{Th}}=3+{\frac{4\times6}{10}}=5.4\,\Omega (5.325)
The final solution is,
i(t)=1.78+(2.29-1.78)\mathrm{e}^{-{\frac{5.4}{2}t}}=1.78+0.51\mathrm{e}^{-2.7t}\,\mathrm{A} (5.326)
