In the Figure shown `i_1=10e^(-2t) A, i_2=4A` and `V_C=3e^(-2t)V`. Determine

a. `i_L` and `V_L` b. `V_(ac), V_(ab), V_(cd)`

Explanation

Charge on the capacitor `q=CV_(c)`
`=2.3e^(-2t)`
`=6e^(-2t)`
Current through capacitor `i_(c)=(dq)/(dt)=12e^(-2t)`
Applying junction rule at point Q
`i_(L)=i_(1)+i_(2)+i_(c)=10e^(-2t)+4-12e^(-2t)`
`=(4-2e^(-2t))`
`=2+2(1-e^(-2t))`

Voltage across inductor
`V_(L)=L(di_(L))/(dt)`
`=4(d)/(dt)(4-2e^(-2t))`
`=16e^(-2t)`
`v_(a)-i_(1)R_(1)+i_(2)R_(2)=v_(c)`
`v_(a)-v_(c)=i_(1)R_(1)-i_(2)R_(2)`
`=10e^(-2t)xx2-4xx3`
`=20e^(-2t)-12`