Answer:
The frequency is [tex]f = 0.221 \ Hz[/tex]
Explanation:
From the question we are told that
The time taken for it to decay to half its original size is [tex]t = 3.40 \ ms = 3.40 *10^{-3} \ s[/tex]
Let the voltage of the capacitor when it is fully charged be [tex]V_o[/tex]
Then the voltage of the capacitor at time t is said to be [tex]V = \frac{V_o}{2}[/tex]
Now this voltage can be mathematical represented as
[tex]V = V_o * e ^{-\frac{t}{RC} }[/tex]
Where RC is the time constant
substituting values
[tex]\frac{V_o}{2} = V_o * e ^{-\frac{3.40 *10^{-3}}{RC} }[/tex]
[tex]0.5 = e^{-\frac{3.40 *10^{-3}}{RC} }[/tex]
[tex]- \frac{0.5}{RC} = ln (0.5)[/tex]
[tex]-\frac{0.5}{RC} = -0.6931[/tex]
[tex]RC = 0.721[/tex]
Generally the cross-over frequency for a low pass filter is mathematically represented as
[tex]f = \frac{1}{2 \pi * RC }[/tex]
substituting values
[tex]f = \frac{1}{2* 3.142 * 0.72 }[/tex]
[tex]f = 0.221 \ Hz[/tex]
