A 10mH Coil is connected in series with a loose free capacitor to a variable frequency source of 20 V . A current in the circuit has a maximum value of 0.2A at a frequency of 100kH. Calculate value of capacitance , the quality factor of the coil and the half power frequencies

To solve this problem, we need to use several formulas related to AC circuits, resonance, and quality factor.

Step 1: Calculate the Resonant Frequency

The resonant frequency (fr) is given by the formula:

fr = 1 / (2π√(LC))

Where L is the inductance and C is the capacitance. We know the resonant frequency (100kHz) and the inductance (10mH), so we can rearrange the formula to solve for C:

C = 1 / (4π²fr²L)

Substituting the given values:

C = 1 / (4π² * (100kHz)² * 10mH) C = 1 / (4 * 3.14² * (100 * 10^3 Hz)² * 10 * 10^-3 H) C = 0.2533 nF

Step 2: Calculate the Quality Factor

The quality factor (Q) of a coil is given by the formula:

Q = fr * L / R

Where R is the resistance. We can find R using Ohm's law (V = IR), rearranged to R = V / I:

R = 20V / 0.2A = 100Ω

Substituting the given values into the formula for Q:

Q = 100kHz * 10mH / 100Ω Q = 10

Step 3: Calculate the Half Power Frequencies

The half power frequencies (f1 and f2) are given by the formulas:

f1 = fr - (fr / (2Q)) f2 = fr + (fr / (2Q))

Substituting the given values:

f1 = 100kHz - (100kHz / (2 * 10)) = 95kHz f2 = 100kHz + (100kHz / (2 * 10)) = 105kHz

So, the capacitance is approximately 0.2533 nF, the quality factor of the coil is 10, and the half power frequencies are 95kHz and 105kHz.