Answer:
The amount of time is 31.45 sec.
Explanation:
Given that,
Voltage = 10.0 V
Resistance [tex]R=2.80\times10^{6}\ \Omega[/tex]
Capacitance [tex]C=3.50\ \mu F[/tex]
We need to calculate the time constant
Using formula of time constant
[tex]\tau=RC[/tex]
Put the value into the formula
[tex]\tau=2.80\times10^{6}\times3.50\times10^{-6}[/tex]
[tex]\tau=9.8\ sec[/tex]
The amount of time required for the current in the circuit to decay to 4.00% of its original value.
We need to calculate the amount of time
Using formula of charge
[tex]Q=Q_{max}(1-e^{\dfrac{-t}{RC}})[/tex]
Put the value into the formula
[tex]96\%Q_{max}=Q_{max}(1-e^{\dfrac{-t}{RC}})[/tex]
[tex]0.96=(1-e^{\dfrac{-t}{\tau}})[/tex]
[tex]0.96=(1-e^{\dfrac{-t}{\tau}})[/tex]
[tex]0.04=e^{\dfrac{t}{\tau}}[/tex]
[tex]\dfrac{-t}{\tau}=ln(0.04)[/tex]
[tex]t=3.21\times9.8[/tex]
[tex]t=31.45\ sec[/tex]
Hence, The amount of time is 31.45 sec.
