Question

A capacitor has a peak current of 310 μa when the peak voltage at 300 khz is 3.0 v .

Answer 1

I think the question for this problem, is to find the capacitance. The working equation would be:

C = Q/V,
where
Q = It

Since f = 1/t, t = (1/300 kHz)(1 kHz/1,000 Hz) = 3.33×10⁻⁶ s
So,
Q = (310×10⁻⁶ A)(3.33×10⁻⁶ s)
Q = 1.033×10⁻⁹ C

Thus,
C = (1.033×10⁻⁹ C)/(3 V)
C = 3.44×10⁻¹⁰ F