Answer :
Answer:
The voltage is [tex]V = 0.993V_b[/tex]
Explanation:
From the question we are told that
The time that has passed is [tex]t = \frac{\tau}{2}[/tex]
Here [tex]\tau[/tex] is know as the time constant
The voltage of the power source is [tex]V_b[/tex]
Generally the voltage equation for charging a capacitor is mathematically represented as
[tex]V = V_b [1 - e^{- \frac{t}{\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{\tau}{2\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{1}{2} }][/tex]
=> [tex]V = 0.993V_b[/tex]