A fully charged capacitor and a 0.32-H inductor are connected to form a complete circuit. If the circuit oscillates with a frequency of 4.1 ✕ 103 Hz, determine the capacitance of the capacitor.
Xc = Xl = 2pi*F*L = 6.28*4100*0.32 = 8244 ohms.
C = 1/(2pi*F*Xc) = 1/(6.28*4100*8244) = 4.7*10^-9 farads = 0.0047 uF.
To determine the capacitance of the capacitor, we can use the formula for the natural frequency of an RLC circuit:
f = 1 / (2π√(LC))
Where:
- f is the frequency of the oscillation
- L is the inductance of the inductor
- C is the capacitance of the capacitor
In this case, we are given the frequency (f = 4.1 × 10^3 Hz) and the inductance (L = 0.32 H). We need to solve for the capacitance (C).
Rearranging the formula, we get:
C = 1 / (4π²Lf²)
Plugging in the values, we have:
C = 1 / (4π² * 0.32 * (4.1 × 10^3)²)
C = 1 / (4 * 3.1416² * 0.32 * (4.1 × 10^3)²)
C ≈ 1 / (4 * 3.1416² * 0.32 * 4.1² * 10^6)
C ≈ 1 / (4 * 9.8696 * 0.32 * 16.81 * 10^6)
C ≈ 1 / (0.32 * 533.3676288)
C ≈ 0.98792 µF
Therefore, the capacitance of the capacitor is approximately 0.98792 microfarads (µF).