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)` and `V_(cd)`

Explanation

Given
`i_(1)=10e^(-2t)A` , `i_(2)=4A` , `V_(c)=3e^(-2t)`
(a) Charge on capacitor
`q=CV_(c)=2xx3e^(-2t)(-2)=-12e^(-2t)`
Current is `-ve` i.e. discharging of capacitor
`|i_(c)|=12e^(-2t)`
`i_(c)+i_(L)=i_(1)+i_(2)`
`12e^(-2t)+i_(L)=10e^(2t)+4`
`i_(L)=4-2e^(-2t)`
`i_(L)` increases with time
`V_(L)=L(di_(L))/(dt)=4(d)/(dt)(4-2e^(-2t))=16e^(-2t)V`
(b) `V_(ac)`=?
`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)V`
`V_(ab)`=?
`V_(a)-i_(1)R_(1)+V_(c)=V_(b)`
`V_(a)-V_(b)=i_(1)R_(1)+V_(C)=10e^(-2t)xx2+3e^(-2t)`
`V_(cd)`=?
`V_(c)-i_(2)R_(2)+V_(L)=V_(d)`
`V_(c)-V_(d)=i_(2)R_(2)+V_(L)=4xx3+16e^(-2t)`
`=(12+16e^(-2t))V`