Question

If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the frequency is closest to:

Answer 1

Answer:

5.19 x 10³Hz

Explanation:

The capacitive reactance, $X_{C}$, which is the opposition given to the flow of current through the capacitor is given by;

$X_C = \frac{1}{2\pi fC }$

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 $X_{C}$             [Substitute the value of

=> V = I x $\frac{1}{2\pi fC}$      [Make f the subject of the formula]

=> f = $\frac{I}{2\pi VC}$                    ---------------------(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 = $\frac{0.00333}{2 * 3.142 * 12.0 * 8.50 * 10^{-9}}$            [Taking $\pi$ = 3.142]

f = 5.19 x 10³Hz

Therefore, the frequency is closest to f = 5.19 x 10³Hz