A 15 $\mu F$ capacitor is connected to a 240V, 50Hz ac. source. Calculate the reactance of the capacitance [$\pi = 3.142$]

Analyze the question: We need the capacitive reactance (Xc) of a 15 μF capacitor on a 50 Hz supply. The 240 V is not needed for this part.

Concept: Capacitive reactance is the opposition a capacitor offers to alternating current:

Xc = 1 / (2πfC)

Solution:

• f = 50 Hz
• C = 15 μF = 15 × 10⁻⁶ F
• π = 3.142

2πfC = 2 × 3.142 × 50 × 15 × 10⁻⁶ = 314.2 × 15 × 10⁻⁶ = 4713 × 10⁻⁶ = 4.713 × 10⁻³

Xc = 1 / (4.713 × 10⁻³) ≈ 212.2 Ω

Answer: Xc ≈ 212 Ω, which is option D.

Tip: Be careful with the units. Always convert μF to F (× 10⁻⁶). A common mistake is to forget the factor of 2 in 2πf. If you leave it out, you get about 106 Ω, which is option C and is placed there as a distractor. Reactance is 1/(ωC), and ω = 2πf.