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Sagot :
Answer:
C = 2.9 10⁻⁵ F = 29 μF
Explanation:
In this exercise we must use that the voltage is
V = i X
i = V/X
where X is the impedance of the system
in this case they ask us to treat the system as an RLC circuit in this case therefore the impedance is
X = [tex]\sqrt{R^2 + ( wL - \frac{1}{wC})^2 }[/tex]
tells us to take inductance L = 0.
The angular velocity is
w = 2π f
the current is required to be half the current at high frequency.
Let's analyze the situation at high frequency (high angular velocity) the capacitive impedance is very small
[tex]\frac{1}{wC}[/tex] →0 when w → ∞
therefore in this frequency regime
X₀ = [tex]\sqrt{R^2 + ( \frac{1}{2\pi 2 10^4 C} )^2 } = R \sqrt{ 1+ \frac{8 \ 10^{-10} }{RC} }[/tex]
the very small fraction for which we can despise it
X₀ = R
to halve the current at f = 200 H, from equation 1 we obtain
X = 2X₀
let's write the two equations of inductance
X₀ = R w → ∞
X= 2X₀ = [tex]\sqrt{R^2 +( \frac{1}{wC} )^2 }[/tex] w = 2π 200
we solve the system
2R = \sqrt{R^2 +( \frac{1}{wC} )^2 }
4 R² = R² + 1 / (wC) ²
1 / (wC) ² = 3 R²
w C = [tex]\frac{1}{\sqrt{3} } \ \frac{1}{R}[/tex]
C = [tex]\frac{1}{\sqrt{3} } \ \frac{1}{wR}[/tex]
let's calculate
C = [tex]\frac{1}{\sqrt{3} } \ \frac{1}{2\pi \ 200 \ 9}[/tex]
C = 2.9 10⁻⁵ F
C = 29 μF
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