Correct Answer - A
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) = (20 e^(-2t) - 12) V`
At `t = 0, V_(AC) = 8V`, At `t = oo, V_(AC) = 12 V`
Therefore, `V_(AC)` decreases exponentially from `8V` to - `- 12V`