In Fig. (a) shown, $i_1=10e^{-2t}\text{ A},i_2=4\text{A}$ and $V_C=3e^{-2t}V$.

Determine: $V_{AB}$ and $V_{CD}$.

We have from $A$ to $B$ 

$V_A-i_1{R}_1+V_C=V_B$

$V_{AB}=V_A-V_B=i_1{R}_1-V_C$

Substituting the values we have, 

$V_{AB}=\left(10e^{-2t}\right)\left(2\right)-3e^{-2t}$

$V_{AB}=17e^{-2t}\text{V}$

Thus, $V_{AB}$ decreases exponentially from $17\text{V }\mathrm{to} 0.$

As we move from $C$ to $D$,

$V_C-i_2{R}_2-i_L=V_D$

Or, $V_{CD}=V_C-V_D=i_2{R}_2+i_L$

Substituting the values we have 

$V_{CD}=\left(4\right)\left(3\right)+16e^{-2t}$

$V_{CD}=(12+16e^{-2t})\text{V}$

At $t=0,V_{CD}=28\text{V}$

And at $t=\infty ,V_{CD}=12\text{V}$

i.e. $V_{CD}$ decreases exponentially from $28\text{V }\mathrm{to}12\text{V}$.