Respuesta :
Explanation:
Given that,
Capacitor [tex]C=1.66\ \mu F[/tex]
Resistor [tex]R=80.0\ \Omega[/tex]
Peak voltage = 5.10 V
(A). We need to calculate the crossover frequency
Using formula of frequency
[tex]f_{c}=\dfrac{1}{2\pi R C}[/tex]
Where, R = resistor
C = capacitor
Put the value into the formula
[tex]f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}[/tex]
[tex]f_{c}=1198.45\ Hz[/tex]
(B). We need to calculate the [tex]V_{R}[/tex] when [tex]f = \dfrac{1}{2f_{c}}[/tex]
Using formula of [tex]V_{R}[/tex]
[tex]V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})[/tex]
Put the value into the formula
[tex]V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})[/tex]
[tex]V_{R}=2.280\ Volt[/tex]
(C). We need to calculate the [tex]V_{R}[/tex] when [tex]f = f_{c}[/tex]
Using formula of [tex]V_{R}[/tex]
[tex]V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})[/tex]
[tex]V_{R}=3.606\ Volt[/tex]
(D). We need to calculate the [tex]V_{R}[/tex] when [tex]f = 2f_{c}[/tex]
Using formula of [tex]V_{R}[/tex]
[tex]V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})[/tex]
[tex]V_{R}=4.561\ Volt[/tex]
Hence, This is the required solution.
