A 30.0μF capacitor is connected to an ac voltage source with an rms voltage of 150V and a frequency of 2.49KHz. What is the capacitive reactance of the circuit? Assume that there is no resistance in the circuit, and answer in units of ohms
Xc = 1/(2pi*F*C) Ohms.
F = 2490 Hz
C = 30*10^-6 Farads.
2pi = 6.28
Xc = ?
To find the capacitive reactance of the circuit, you can use the formula:
Xc = 1 / (2πfC)
where Xc is the capacitive reactance, f is the frequency, and C is the capacitance.
In this case, the given values are:
C = 30.0μF = 30.0 × 10^(-6) F
f = 2.49KHz = 2.49 × 10^3 Hz
Substituting these values into the formula, we get:
Xc = 1 / (2π × 2.49 × 10^3 × 30.0 × 10^(-6))
Now let's calculate it step by step:
1. Convert the capacitance to Farads:
30.0μF = 30.0 × 10^(-6) F
2. Plug the values into the formula:
Xc = 1 / (2π × 2.49 × 10^3 × 30.0 × 10^(-6))
3. Simplify the expression:
Xc = 1 / (2π × 2.49 × 10^3 × 30.0 × 10^(-6))
Xc = 1 / (2 × 3.14159... × 2.49 × 10^3 × 30.0 × 10^(-6))
Xc = 1 / (6.28318... × 2.49 × 10^3 × 30.0 × 10^(-6))
Xc ≈ 1 / (4.93646... × 10^(-1))
Xc ≈ 2.02503... Ω
Therefore, the capacitive reactance of the circuit is approximately 2.03 ohms (Ω).