正确答案：
http://sc0.ykt.io/ue_i/20191113/1194455928540368896.png、SolutionStep 1: Solving ideaAfter S is switched on, the circuit can be divided into two first-order circuits. (Notice: not superposition) By calculating ${i_1}$ in circuit (a) and ${i_L}$ in circuit (b), then we get $i = {i_1} - {i_L}$.Step 2: Solve ${u_C}({0^ - })$ and ${i_L}({0^ - })$ in ${0^ - }$ circuit. At this time, S is not switched on and the circuit is in steady state.${i_L}({0^ - }) = frac{{10}}{{10 + 5 + 5}} = 0.5{rm{A}}$${u_C}({0^ - }) = {i_L}({0^ - }) times (5 + 5) = 5{rm{V}}$Step 3: Solve ${i_{rm{1}}}left( t right)$ in circuit (a) with three elements method${u_C}({0^ + }) = {u_C}({0^ - }) = 5{rm{V}}$${i_1}({0^ + }) = frac{{{u_C}({0^ + }) - 5}}{5} = 0{rm{A}}$${i_1}(infty ) = frac{{10 - 5}}{{10 + 5}} = 0.333{rm{A}}$${tau _1} = (10//5) times 1 = frac{{10}}{3}{rm{s}}$Then we get ${i_1}(t) = 0.333 - 0.333{{rm{e}}^{ - 0.3t}}{rm{A}}quad t ge {0^ + }$.Step 4: Solve ${i_L}left( t right)$ in circuit (b) with three elements method${i_L}({0^ + }) = {i_L}({0^ - }) = 0.5{rm{A}}$${i_L}(infty ) = frac{5}{5} = 1{rm{A}}$${tau _2} = frac{2}{5} = 0.4{rm{s}}$Then we get ${i_L}(t) = 1 + (0.5 - 1){{rm{e}}^{ - 2.5t}} = 1 - 0.5{{rm{e}}^{ - 2.5t}}{rm{A}}quad t ge 0$.Step 5:$i(t) = {i_1}(t) - {i_L}(t) = - 0.667 - 0.333{{rm{e}}^{ - 0.3t}} + 0.5{{rm{e}}^{ - 2.5t}}{rm{A}}quad t ge {0^ + }$
