A solenoid has an inductance of 10 henry and a resistance of 2 ohm. It is connected to a 10 volt battery. How long will it take for the magnetic energy to reach 1/4 of its maximum value?

Explanation

The L-R circuit is shown in the figure . Let the current through the circuit be I after a time t. Then ` L ""(di)/(dt)+ Ri=V` or `(di)/(V-Ri) =(dt)/(L)`
Integrating this expression and keeping i=0 when t =0 we get `I =(V)/(R ) (1-e^(R//L))=i_(0)(1-e^(-Rt//L)),` where `i_(0) = V//R ` maximum current Magnetic energy stored =`(1)/(2) Li^(2)`
Here `,(1)/(2) Li^(2)=(1)/(4) ((1)/(2)Li_(0)^(2)) ` or `i=(i_(0))/(2)=(V)/(2R)`
Substituting the values in equation (1) we get
`(V)/(2R)=(V)/(R) (1-e^(2t//10))"or" 0.5=1-e^(-0.2t)`
or `e^(-02t)=0.5,` which gives t = 3.478 sec .