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)`

Options

A. `8e^(-2t)V`
B. `(1)/(2) e^(-3t) V`
C. `17 e^(-2t)V`
D. `16 e^(-2t)`V`

Answer

C

Explanation

Instanctaneous current in the capacitor,
`q = CV_(C) = (2)(3e^(-2t)) = 6e^(-2t) A`
Current, `i_(C) = (dq)/(dt) = - 12e^(-2t)A`
This current flows from `B` to `O`.
From `KVL`, we have
`i_(L) = i_(1) + i_(2) + i_(C) = 10e^(-2t) + 4 - 12e^(-2t)`
`= (4 - 2t^(-2t)) A = [2 + 2(1 - e^(-2t))]A`
`i_(L) vs`. time graph is as shows in Fig lt
`i_(L)` increases from `2 A` to `4 A` expontially.

`V_(L) = L(di_(L))/(dt)`
`= (4)(d)/(dt) (4 - 2e^(-2t)) = 16e^(-2t) V`
`V_(L)` decreases exponentially from `16 A` to `0` as shows in Fig
To determine `V_(AC)`, we begin from `A` and at `C`. From `KVL`,
we have

`V_(A) - i_(1)R_(1) + i_(2)R_(2) = V_(C)`
`V_(A) - V_(C) = i_(1)R_(1) - i_(2)R_(2)`
Substituting the values, we have
`V_(AC) = (10 e^(-2t))(2) - (4)(3)`

`V_(AC) = (20e^(-2t) - 12) V`
At `t = 0`, `V_(AC) = 8V`
At `t = oo`, `V_(AC) = - 12 V`
Therefore, `V_(AC)` decrease exponentially from `8 V` to `12 V`.
Similarly, 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)=^((10e^(-2t)))(2)-3e^(-2t)`
`V_(AB) =^(17e^(-2t))V`
Thus, `V_(AB)` decreases exponentially from `17 V` to `0`.
As we move from `C` to `D`,
`V_(C) - i_(2)R_(2) - V_(L) = V_(D)`
`V_(CD)= V_(C) - V_(D) = i_(2)R_(2) + V_(L)`
Substituting the values we have,
`V_(CD)= (4)(3) + 16e^(-2t)`

`V_(AD) = (12 + 16e^(-2t))V`
At `t = 0, V_(CD) = 28 V`
and at `t = oo, V_(CD) = 12 V`
i.e., `V_(CD)` decreases exponentially from `28 V` to `12 V`.