5.19 x 10³Hz
The capacitive reactance, [tex]X_{C}[/tex], which is the opposition given to the flow of current through the capacitor is given by;
[tex]X_C = \frac{1}{2\pi fC }[/tex]
Where;
f = frequency of the signal through the capacitor
C = capacitance of the capacitor.
Also, from Ohm's law, the voltage(V) across the capacitor is given by the product of current(I) and the capacitive reactance. i.e;
V = I x [tex]X_{C}[/tex] [Substitute the value of
=> V = I x [tex]\frac{1}{2\pi fC}[/tex] [Make f the subject of the formula]
=> f = [tex]\frac{I}{2\pi VC}[/tex] ---------------------(i)
From the question;
I = 3.33mA = 0.00333A
C = 8.50nF = 8.50 x 10⁻⁹F
V = 12.0V
Substitute these values into equation (i) as follows;
f = [tex]\frac{0.00333}{2 * 3.142 * 12.0 * 8.50 * 10^{-9}}[/tex] [Taking [tex]\pi[/tex] = 3.142]
f = 5.19 x 10³Hz
Therefore, the frequency is closest to f = 5.19 x 10³Hz
