A 33.0-mH inductor has a reactance of 2.20 kΩ.

(a) What is the frequency of the ac current that passes through the inductor?


(b) What is the capacitance of a capacitor that has the same reactance at this frequency?


(c) The frequency is tripled, so that the reactances of the inductor and capacitor are no longer equal. What is the new reactance of the inductor?


(d) What is the new reactance of the capacitor?

See previous post.

To answer these questions, we need to understand the formulas related to inductance, reactance, and frequency.

(a) The reactance of an inductor is given by the formula Xl = 2πfL, where Xl is the reactance, f is the frequency, and L is the inductance. Rearranging the formula, we have f = Xl / (2πL). Plugging in the given values Xl = 2.20 kΩ and L = 33.0 mH (which is equivalent to 0.033 H), we can calculate the frequency:

f = (2.20 kΩ) / (2π * 0.033 H) = 33.67 kHz

So, the frequency of the AC current passing through the inductor is approximately 33.67 kHz.

(b) The formula for the reactance of a capacitor is Xc = 1 / (2πfC), where Xc is the reactance, f is the frequency, and C is the capacitance. We need to find the capacitance that has the same reactance as the inductor at the given frequency. Rearranging the formula, we have C = 1 / (2πfXc). Plugging in the given values f = 33.67 kHz and Xc = 2.20 kΩ, we can calculate the capacitance:

C = 1 / (2π * 33.67 kHz * 2.20 kΩ) = 2.28 nF

So, the capacitance of a capacitor that has the same reactance at this frequency is approximately 2.28 nF.

(c) When the frequency is tripled, we need to find the new reactance of the inductor. Using the same formula as before, Xl = 2πfL, we can calculate the new reactance. Plugging in the new frequency (which is 3 times the initial frequency), f = 3 * 33.67 kHz, and the same inductance L = 0.033 H, we have:

Xl = 2π * (3 * 33.67 kHz) * 0.033 H = 6.68 kΩ

So, the new reactance of the inductor is 6.68 kΩ.

(d) Similarly, we need to find the new reactance of the capacitor when the frequency is tripled. Using the same formula as before, Xc = 1 / (2πfC), we can calculate the new reactance. Plugging in the new frequency, f = 3 * 33.67 kHz, and the same capacitance C = 2.28 nF, we have:

Xc = 1 / (2π * (3 * 33.67 kHz) * 2.28 nF) ≈ 0.555 kΩ

So, the new reactance of the capacitor is approximately 0.555 kΩ.