Answer:
The fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is
[tex]k = 0.903[/tex]
Explanation:
From the question we are told that
The time constant [tex]\tau = 3[/tex]
The potential across the capacitor can be mathematically represented as
[tex]V = V_o (1 - e^{- \tau})[/tex]
Where [tex]V_o[/tex] is the voltage of the capacitor when it is fully charged
So at [tex]\tau = 3[/tex]
[tex]V = V_o (1 - e^{- 3})[/tex]
[tex]V = 0.950213 V_o[/tex]
Generally energy stored in a capacitor is mathematically represented as
[tex]E = \frac{1}{2 } * C * V ^2[/tex]
In this equation the energy stored is directly proportional to the the square of the potential across the capacitor
Now since capacitance is constant at [tex]\tau = 3[/tex]
The energy stored can be evaluated at as
[tex]V^2 = (0.950213 V_o )^2[/tex]
[tex]V^2 = 0.903 V_o ^2[/tex]
Hence the fraction of the energy stored in an initially uncharged capacitor is
[tex]k = 0.903[/tex]
