Question

At `t=0` , switch `S` closed. Find the ratio of the rate at which magnetic energy stored in the coil to the rate at which energy is supplied by the battery at time `t=tau=L//R` .

Answer 1

In `R-L` circuit, growth of current
`i=i_(0)(1-e^(-t//tau))`
where `i_(0)=(E)/(R)` , `tau(L)/(R)`
`(di)/(dt)=(i_(0)e^(-t//tau))/(tau)`
Magnetic energy stored in the coil
`U=(1)/(2)Li^(2)`
`alpha(dU)/(dt)=(1)/(2)L.2i(di)/(dt)=Li(di)/(dt)`
The rate at which energy supplied by battery
`beta=Ei`
`(alpha)(beta)=(Lidi//dt)/(Ei)=(L)/(E)(di)/(dt)`
`(L)/(E).(i_(0)e^(-t//tau))/(tau)`
At `t=tau`
`(alpha)/(beta)=(L)/(E)(i_(0)e^(-1))/(tau)=(L)/(E).(E)/(R).(R)/(L.e)`
`(1)/(e)`