Define terminal velocity and find an expression for it.

Explanation

Mean value of alternating current : Mean value of a.c. over any half cycle is that steady current which sends the same charge through a circuit in half time period of alternating current as is sent by the alternating current in the same time through the same circuit.
Mean value of a.c. is denoted by `I_(av)" or "I_m` , At any time, alternating current flowing through the circuit is given by
`I=I_0 sin omegat " "...(i)`
Let dq = Small amount of charge sent by a.c. in small time dt.
Suppose I remains same constant for time dt
`dq = I dt`
`dq = I_0 sin omegat dt` [Using (1)]
Charge sent by a.c. in half cycle is given by
`intdq=int_(0)^(T//2)I_0sinomegatdt `
`q=I_0int_(0)^(T//2)sinomegatdt=I_0|(-cosomegat)/omega|_0^(T//2)`
`q=-I_0/omega|cos""(omegaT)/2-cos0^0|`
`q=-I_0/omega|cos""(2pi)/T(T/2)-1|`
`q=-I_0/omega|-1-1|=(2I_0)/omega`
`q=|(2I_0)/(2pi//T)|=(I_0T)/(pi) " "...(ii)`
Let `I_(av)` = Mean value of a.c. over the half cycle of a.c.
Then charge sent by `I_(av)` in the time T/2 is given by
`q=I_(av)(T/2)" "...(iii)`
From eqns. (ii) and (iii)
`I_(av)(T/2)=(I_0T)/pi`
`implies I_(av)=(2I_0)/pi=0.637I_0`
Hence mean value of a.c is 0.637 time the peak value of a.c. i.e. 63.7% of the peak value of a.c.