A 3.0 uF capacitor has 1.0 kQ reactance. (a) What's the frequency of the applied voltage? (b) What inductance would give the same reactance at this frequency? (c) How would the reactances compare if the frequency were doubled? (1 point)

(a) To find the frequency of the applied voltage, we use the formula for capacitive reactance:

$X_C = \frac{1}{2 \pi f C}$

Given: $X_C = 1.0 \, \text{k}\Omega = 1000 \, \Omega$ $C = 3.0 \, \mu\text{F} = 3.0 \times 10^{-6} \, \text{F}$

Rearranging the formula to solve for frequency $f$:

$f = \frac{1}{2 \pi X_C C}$

Substitute the given values:

$f = \frac{1}{2 \pi \times 1000 \, \Omega \times 3.0 \times 10^{-6} \, \text{F}}$

$f = \frac{1}{2 \pi \times 3.0 \times 10^{-3}}$

$f = \frac{1}{6 \pi \times 10^{-3}}$

$f \approx \frac{1}{0.01885}$

$f \approx 53.05 \, \text{Hz}$

(b) To find the inductance that would give the same reactance at this frequency, we use the formula for inductive reactance:

$X_L = 2 \pi f L$

Given: $X_L = 1.0 \, \text{k}\Omega = 1000 \, \Omega$ $f = 53.05 \, \text{Hz}$

Rearranging the formula to solve for inductance $L$:

$L = \frac{X_L}{2 \pi f}$

Substitute the given values:

$L = \frac{1000 \, \Omega}{2 \pi \times 53.05 \, \text{Hz}}$

$L = \frac{1000}{333.14}$

$L \approx 3.0 \, \text{H}$

(c) If the frequency were doubled, the new frequency $f'$ would be:

$f' = 2 \times 53.05 \, \text{Hz} = 106.1 \, \text{Hz}$

For the capacitive reactance $X_C'$:

$X_C' = \frac{1}{2 \pi f' C}$

$X_C' = \frac{1}{2 \pi \times 106.1 \, \text{Hz} \times 3.0 \times 10^{-6} \, \text{F}}$

$X_C' = \frac{1}{2 \pi \times 3.183 \times 10^{-4}}$

$X_C' \approx \frac{1}{0.002}$

$X_C' \approx 500 \, \Omega$

For the inductive reactance $X_L'$:

$X_L' = 2 \pi f' L$

$X_L' = 2 \pi \times 106.1 \, \text{Hz} \times 3.0 \, \text{H}$

$X_L' = 2 \pi \times 318.3$

$X_L' \approx 2000 \, \Omega$

So, if the frequency were doubled, the capacitive reactance would be halved to approximately 500 Ω, and the inductive reactance would be doubled to approximately 2000 Ω.