The frequency of a LC circuit is 125kHz. If the capacitor filled with a dielectric the frequency decreases by 25kHz. Find the dielectric constant.

To find the dielectric constant of the capacitor in an LC circuit, we first need to understand how the frequency of the circuit is affected by the presence of a dielectric material. The frequency of an LC circuit is given by the formula:

Frequency of an LC Circuit

The resonant frequency (f) of an LC circuit is determined by the inductance (L) and capacitance (C) using the formula:

f = 1 / (2π√(LC))

In this scenario, the initial frequency of the circuit is 125 kHz, and when a dielectric is introduced into the capacitor, the frequency decreases to 100 kHz (125 kHz - 25 kHz). The presence of a dielectric affects the capacitance of the capacitor, which in turn affects the frequency of the circuit.

Capacitance with Dielectric

The capacitance of a capacitor filled with a dielectric material is given by:

C' = k * C

where:

• C' is the capacitance with the dielectric
• C is the original capacitance without the dielectric
• k is the dielectric constant of the material

Relating Frequencies and Capacitance

Since the frequency decreases when the dielectric is introduced, we can set up the relationship between the two frequencies:

f = 1 / (2π√(L * C)) and f' = 1 / (2π√(L * C'))

Substituting the expression for C' into the frequency formula gives:

f' = 1 / (2π√(L * k * C))

Setting Up the Equation

Now we can relate the two frequencies:

f' = f / √k

Plugging in the values we have:

100 kHz = 125 kHz / √k

Solving for the Dielectric Constant

Rearranging the equation to solve for k gives:

√k = 125 kHz / 100 kHz

√k = 1.25

Now, squaring both sides to find k:

k = (1.25)² = 1.5625

Final Result

The dielectric constant of the material used in the capacitor is approximately 1.56. This means that the dielectric material increases the capacitance by about 56% compared to the original capacitance without the dielectric.