Self - inductance `0.8xx10^(-4)`H of a uniformly wound solenoid, having resistance 3 `Omega` is broken up into two indentical coils. Those coils are connected in series across a 6 V battery of negligible resistance. Find time constant and steady state current.

Explanation

For a solenoid, self inductance is given by
`L=(mu_(0)N^(2)pir^(2))/(l)`
where, r = radius of colenoid, l = length of solenoid,
N = number of turns
are we know that, resistance is given by
`R=rho(l)/(A)=rho(N2pir)/(A)`
So, on breaking it into two parts, now inductance and resistance is given as
`N'=(N)/(2),l'=(l)/(2)rArrL'=(mu_(0)(N//2)^(2)pir^(2))/(l//2)=(L)/(2)`
`R'=(rho(N//2)xx2pir)/(A)=R`

So, the equivalent inductance, `L_(eq)=L_(1)+L_(2)`
`=L//2+L//2=L`
and eqivalent resistance, `R_(eq)=R_(1)+R_(2)=R//2+R//2=R`
So, the time constant,
`tau=(L_(eq))/(R_(eq))=(0.8xx10^(-4))/(3)=0.267xx10^(-4)s`
and current, `I=(E)/(R_(eq))=(E)/(R)=(6)/(3)=2A`