Question

The capacitive reactance offered by a capacitance of 10 microfarads to an alternating current of \( i = I_m \sin(100\pi t - \frac{\pi}{6}) \) amperes is

• A. 318.31 ohms
• B. 0.31831 ohm
• C. 31.831 ohms
• D. 3183.1 ohms

Hint:

To calculate capacitive reactance, use the formula \( X_C = \frac{1}{2\pi f C} \), where \( f \) is the frequency and \( C \) is the capacitance.

Answer 1

The Correct Option is A

Step 1: Read the angular frequency directly from the current expression: \( \omega = 100\pi \, \text{rad/s} \).
Step 2: Apply \( X_C = \dfrac{1}{\omega C} = \dfrac{1}{100\pi \times 10\times10^{-6}} \).
Step 3: Evaluating, \( 100\pi \times 10\times10^{-6} \approx 3.1416\times10^{-3} \), so \( X_C \approx 318.31 \, \Omega \).
\[ \boxed{X_C \approx 318.31 \, \Omega} \]

Answer 2

Before computing precisely, we can narrow down the answer using order-of-magnitude reasoning. A 10 microfarad capacitor at a typical power-line-like frequency (\( \omega = 100\pi \approx 314 \, \text{rad/s} \), i.e. \( f = 50 \, \text{Hz} \)) is known from common circuit experience to have a reactance in the low hundreds of ohms range — neither a fraction of an ohm nor thousands of ohms. Let's check each option against this order-of-magnitude expectation.

1. 318.31 ohms: This falls squarely in the expected "low hundreds of ohms" range for a 10 µF capacitor at 50 Hz.
2. 0.31831 ohm: This is far too small — such a low reactance would require either a very large capacitance or an extremely high frequency, neither of which applies here.
3. 31.831 ohms: This is roughly ten times too small for the given capacitance and frequency combination.
4. 3183.1 ohms: This is roughly ten times too large, which would instead correspond to a capacitance about 10 times smaller (around 1 µF) at the same frequency.

The order-of-magnitude check, together with the precise formula, confirms the capacitive reactance is 318.31 ohms.

Therefore, the correct answer is 318.31 ohms.