Saved Bookmarks
| 1. |
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.A. `(E^(2))/(4R)`B. `(E^(2))/(R )`C. `(4E^(2))/( R)`D. `(2E^(2))/(R )` |
|
Answer» Correct Answer - A 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)` |
|