An inductance L and a resistance R are connected in series with a battery of emf e. Find the maximum rate at which the energy is stored in the magnetic field.

Options

A. `(E^(2))/(4R)`
B. `(E^(2))/(R )`
C. `(4E^(2))/( R)`
D. `(2E^(2))/(R )`

Answer

A

Explanation

The graph of current is given by :
`i=i_(0)(1-e^(-1//tau)) rArr (di)/(dt)=(i_(0))/(tau)e^(i//tau)`
Energy stored in the form of magnetic field energy is:
`U_(B)=(1)/(2)Li^(2)`
`therefore` Rate of increase of magnetic field energy is :
`R=(dU_(B))/(dt)=Li(di)/(dt)=(Li_(0)^(2))/(tau)(1-e^(-1//tau))e^(-1//tau)`
This wiill be maximum when `(dR)/(dt)=0`
`rArr=e^(-1..tau)=1//2`
`R_("max")=(Li_(0)^(2))/)tau)=(Li_(0)^(2))/(tau)(1-(1)/(2))((1)/(2))=(Li(0)^(2))/(4tau)`
`=[(L(E//R)^(2))/(4(L//R))]=(E^(2))/(4R)`