In the circuit shown in figure, E = 10 V, r = `1Omega`,
R = `4Omega` and L = 5H. The circuit is closed at time t = 0.
Then, match the following two columns.

`{:(,"Column I",,"Column II"),(A.,V_(AD)" at t = 0",p,4 V),(B.,V_(BC)" at t = 0",q,8 V),(C.,V_(AB)" at t ="oo,r.,0V),(D.,V_(AD)" at t"=oo,s.,10V):}`

Answer

`(Ararrs;Brarrs;Crarrq;Drarrp)`

Explanation

`V_(AD)` at = 0 = E = 10V
`V_(BC)` at t = 0 E = 10 V
`V_(AB)` at `t = oo`
`I=(E)/(R+r)=(10)/(5)=2A`
`therefore" "V_(AB)=V_(A)-V_(B)=IxxR`
`=2xx4=8V` ltbr. `V_(AD)` at `t = oo`
`V_(AD) = V_(A)-V_(D)=E-Ir`
`=10-(2xx1)=8V`