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

Options

A. (a) `(E^(2))/(4R)`
B. (b) `(E^(2))/(R)`
C. ( c) `(4E^(2))/(R)`
D. (d) `(2E^(2))/(R)`

Answer

A

Explanation

(a) The current is given by
`I = i_(0)(1 - e^(-1//tau)) `rarr` (di)/(dt) = (i_(0))/(tau)e^(-t//tau)`
Energy stored in the form of magnetic field energy is
`U_(B) = (1)/(2)Li^(2)`
`:.` Rate of increase of magneitc field energy is
`R = (dU_(B))/(dt) = Li(di)/(dt) = (Li_(0)^(2))/(tau)(1 - e^(-t//tau))e^(-t//tau)`
This will bemaximum when `(dR)/(dt) = 0`
`rarr` ` e^(-t//tau) = 1//2`
Subsitituting,
`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)`