Answer:
time for maximum energy is 1.718 sec
Explanation:
Maximum energy on the capacitor is given as [tex]=\frac{q_{max}^2}{ 2C}[/tex]
Maximum energy [tex]= \frac{(1/6) Q_2}{2C}[/tex]
The maximum charge is given by
[tex]q_{max} =Qe^{-Rt/2L}[/tex]
Or[tex] \frac{q_{max}}{Q} = e^{-Rt/2L}[/tex]
Or[tex] ln\frac{q_{max}}{Q} = \frac{-Rt}{2L}[/tex]
solving for t
[tex]t = \frac{2L}{R}\frac{1}{2} ln( 2)[/tex]
Putting all value to get desired value
[tex] t = ln(2)\times \frac{L}{R}[/tex]
[tex] t = 0.693\times \frac{(15.2}{6.13} = 1.71 sec[/tex]
