Question

(a) At what frequency would a $\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mH}$ inductor and a$\mathbf{10}\mathbf{}\mathbf{mF}$capacitor have the same reactance? (b) What would the reactance be? (c) Show that this frequency would be the natural frequency of an oscillating circuit with the same Land C.

Short answer

1. The frequency at which the inductance and the capacitance will have the same reactance is $6.5\times {10}^2\mathrm{Hz}$.
2. The reactance would be $24\mathrm{\Omega }$.
3. This frequency is the same as the natural frequency of an oscillating circuit with the same L and C.

Step-by-step solution

The given data

1. Inductance$L=6.0\mathrm{mH}$
2. Capacitance$C=10\mathrm{\mu F}$

Understanding the concept of frequency of LC circuit

When the natural frequency of oscillations and the driving frequency become equal, a phenomenon name resonance takes place. In this condition, the inductive and the capacitive reactance become equal. We equate the equations of inductive reactance and capacitive reactance and solve for angular frequency. Substituting this angular frequency in the formula of frequency, we get the required frequency. Using the angular frequency formula in inductive reactance, we can calculate the inductive reactance.

The inductive reactance of the inductor,

$X_L={\omega }_{d}L$……(i)

The capacitive reactance of the capacitor,

$X_c=\frac{1}{{\omega }_{d}C}$….. (ii)

The relation between frequency and angular frequency,

${\omega }_d=2{\mathrm{\pi f}}_{\mathrm{d}}$…..(iii)

Here, $L$is the inductance of the inductor and$C$ is the capacitance of the capacitor

a) Calculation of the frequency

The two reactance are equal. So, using equations (i) and (ii), we get the angular frequency of the oscillation as:

$$\begin{gathered} X_L=X_C \\ {\omega }_{d}L=\frac{1}{{\omega }_{d}C} \\ {{\omega }^2}_d=\frac{1}{LC} \\ {\omega }_d=\sqrt{\frac{1}{LC}}\left(\mathrm{i}\right) \end{gathered}$$

Thus, the frequency of the oscillation can be given using the above value of equation (1) in equation (iii) as follows:

$$\begin{gathered} f_d=\frac{1}{2\mathrm{\pi }\sqrt{\mathrm{LC}}} \\ {f}_d=\frac{1}{2\mathrm{\pi }\sqrt{\left(6\times {10}^{-3}\mathrm{H}\right)\left(10\times {10}^{-6}\mathrm{F}\right)}} \\ {f}_d=6.5\times {10}^2\mathrm{Hz} \end{gathered}$$

Hence, the value of the frequency is $6.5\times {10}^2\mathrm{Hz}$.

Calculation of the reactance

As both the reactance are same. Thus, the value of the reactance can be given using equation (iii) in equation (i) as follows:

$$\begin{gathered} X_L=2{\mathrm{\pi f}}_{\mathrm{d}}\mathrm{L} \\ {\mathrm{X}}_{\mathrm{L}}=2\mathrm{\pi }\left(6.5\times {10}^2\mathrm{Hz}\right)\left(6\times {10}^{-3}\mathrm{H}\right) \\ {\mathrm{X}}_{\mathrm{L}}=24\mathrm{\Omega } \end{gathered}$$

Hence, the value of the reactance is $24\mathrm{\Omega }$.

c) Calculation of the natural frequency

We know that the natural frequency is given as:

$f=\frac{1}{2\mathrm{\pi }\sqrt{\mathrm{LC}}}$

This is the same as calculated in (a).

So, we can say that the calculated frequency is the natural frequency with the same L and C.