Step 1: Understanding the Concept:
Capacitive reactance (\(X_{C}\)) is the measure of the opposition that a capacitor offers to the flow of alternating current.
Unlike a standard resistor which opposes current by dissipating energy as heat, a capacitor opposes current by storing and releasing energy in its electric field.
In a DC circuit, a capacitor acts as an open circuit (infinite resistance) once it is fully charged because the frequency of DC is zero.
However, in an AC circuit, the direction of current changes continuously, causing the capacitor to charge and discharge repeatedly.
This creates an "effective resistance" that depends on both the physical size of the capacitor (capacitance) and how fast the current changes (frequency).
Higher frequency means the capacitor charges and discharges more quickly, offering less opposition to the current.
Similarly, a larger capacitor can store more charge, which also reduces the opposition to the flow of alternating current.
Step 2: Key Formula or Approach:
The formula used to calculate capacitive reactance is:
\[ X_{C} = \frac{1}{2\pi f C} \]
Where:
\(X_{C}\) = Capacitive Reactance (measured in Ohms, \(\Omega\))
\(f\) = Frequency of the AC source (measured in Hertz, Hz)
\(C\) = Capacitance (measured in Farads, F)
\(\pi \approx 3.14159\)
Step 3: Detailed Explanation:
First, identify the given parameters from the question:
Capacitance, \(C = 50 \text{ \(\mu\)F} = 50 \times 10^{-6} \text{ F} \).
Frequency, \(f = 50 \text{ Hz} \).
Now, substitute these values into our reactance formula:
\[ X_{C} = \frac{1}{2 \times 3.14 \times 50 \times (50 \times 10^{-6})} \]
Let's simplify the denominator calculation:
Multiply the constants: \(2 \times 50 = 100 \).
Multiply by the capacitance value: \(100 \times 50 = 5000 \).
The expression now looks like this:
\[ X_{C} = \frac{1}{3.14 \times 5000 \times 10^{-6}} \]
Convert 5000 to scientific notation: \(5000 = 5 \times 10^{3}\).
\[ X_{C} = \frac{1}{3.14 \times 5 \times 10^{3} \times 10^{-6}} \]
Combine the powers of ten: \(10^{3} \times 10^{-6} = 10^{-3}\).
\[ X_{C} = \frac{1}{15.7 \times 10^{-3}} \]
Move the power of ten to the numerator:
\[ X_{C} = \frac{1000}{15.7} \]
By performing the division:
\[ X_{C} \approx 63.694 \text{ \(\Omega\)} \]
Rounding to one decimal place, we get approximately 63.7 \(\Omega\).
This value tells us that at a frequency of 50 Hz, this specific capacitor behaves like a 63.7-ohm resistor in the circuit.
Step 4: Final Answer:
The capacitive reactance is approximately 63.7 \(\Omega\).
Therefore, the correct option is (B).