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------------------------------------------------ A [tex]25 \mu F[/tex] capacitor is operated at 150 Hz. What is the capacitive reactance?

A. [tex]0.024 \Omega[/tex]
B. [tex]37.5 \Omega[/tex]
C. [tex]450 \Omega[/tex]
D. [tex]42.44 \Omega[/tex]

To find the capacitive reactance, we start with the formula

[tex]$$
X_C = \frac{1}{2 \pi f C},
$$[/tex]

where
- [tex]$f$[/tex] is the frequency in hertz, and
- [tex]$C$[/tex] is the capacitance in farads.

Step 1. Convert the capacitance to farads.
The capacitor is given as [tex]$25\,\mu F$[/tex]. Since

[tex]$$
1\,\mu F = 10^{-6}\,F,
$$[/tex]

we have

[tex]$$
C = 25 \times 10^{-6}\,F = 2.5 \times 10^{-5}\,F.
$$[/tex]

Step 2. Compute the angular frequency.
The angular frequency is

[tex]$$
\omega = 2 \pi f.
$$[/tex]

With [tex]$f = 150\,Hz$[/tex], this becomes

[tex]$$
\omega = 2 \pi (150) \approx 942.48\, \text{rad/s}.
$$[/tex]

Step 3. Compute the product [tex]$2 \pi f C$[/tex].
Multiply the angular frequency by the capacitance:

[tex]$$
2 \pi f C \approx 942.48 \times 2.5 \times 10^{-5} \approx 0.02356.
$$[/tex]

Step 4. Calculate the capacitive reactance.
Now, substitute the computed value into the formula:

[tex]$$
X_C = \frac{1}{0.02356} \approx 42.44\,\Omega.
$$[/tex]

Thus, the capacitive reactance is approximately

[tex]$$
\boxed{42.44\,\Omega}.
$$[/tex]