In the circuit shown, charge `q` varies with time `'t'` as `q=t^(3)-t^(2)+4t`, where `q` is in coulombs and time `'t'` is in second. Find `V_(AB)(V_(A)-V_(B))` in volts at time `t=1` sec

Answer

7

Explanation

`q=t^(3)-t^(2)+4t`
`i=(dq)/(dt)=3t^(2)-2t+4`
`i(t=1)=5` amp
`(di)/(dt)=6t-2`
`(di)/(dt)(t=1)=4`
Current is increasing with time
`q(t=1)=1-1+4=4` coulombs
`V_(A)-V_(B)=4/2+4/2+5xx3/5=7` Volts