Question

A15µF capacitor is connected to a 220V, 50Hz a.c. supply. Its capacitive reactance is *

No scam ans....either I will report the ans​

Answer 1

$\bigstar$ Given

A 15 µF capacitor is connected to a 220 V, 50 Hz A.C. supply

$\bigstar$ To Find

Capacitive reactance

$\bigstar$ Solution

We know that

$\red{\rm \longrightarrow X_{C}=\dfrac{1}{2 \pi fC}}$

Here

• $\rm X_{C}$ = capacitive reactance
• f = frequency
• C = Capacitance

Units

• $\rm X_{C}$ = Ohm (Ω)
• f = hertz (Hz)
• C = Farad (F)

According to the question

We are asked to find the capacitive reactance

Therefore

We must find $\rm X_{C}$

Given that

A 15 µF capacitor is connected to a 220 V, 50 Hz A.C. supply

Hence

• C = 15 µF = 15 x 10⁻⁶ F
• f = 50 Hz

Substituting the values

We get

$\rm \longrightarrow X_{C}=\dfrac{1}{2 \pi\times 50 \times 15 \times 10^{-6}} \ \Omega$

$\rm \longrightarrow X_{C}=\dfrac{1}{100 \pi \times 15 \times 10^{-6}} \ \Omega$

$\rm \longrightarrow X_{C}=\dfrac{1}{1500 \pi \times 10^{-6}} \ \Omega$

We know that

• $\rm \pi$ = 3.14

Substituting the values

We get

$\rm \longrightarrow X_{C}=\dfrac{1}{1500 \times 3.14 \times 10^{-6}} \ \Omega$

$\rm \longrightarrow X_{C}=\dfrac{1}{4710 \times 10^{-6}} \ \Omega$

$\rm \longrightarrow X_{C}=\dfrac{1}{4710} \times 10^{6} \ \Omega$

On further simplification

We get

$\rm \longrightarrow X_{C}=0.0002123 \times 10^{6} \ \Omega$

Hence

$\pink{\rm \longrightarrow X_{C}=212.3 \ \Omega}$

Therefore

$\star$ Capacitive Reactance = 212.3 Ω

Hence

$\checkmark$ Option (c) is correct

Answer 2

Answer:

Option C 212.3 ohm is correct answer.