Question

The capacitor in Figure\(23.55\)(b) will filter high-frequency signals by shorting them to earth/ground. (a) What capacitance is needed to produce a reactance of\(10.0{\rm{ }}m\Omega \)for a\(5.00{\rm{ }}kHz\)signal? (b) What would its reactance be at\(3.00{\rm{ }}Hz\)? (c) Discuss the implications of your answers to (a) and (b).

Short answer

(a.) The capacitance is obtained as: \(3.18{\rm{ }}mF\).

(b.) The reactance is obtained as: \(16.7{\rm{ }}\Omega \).

(c.)The high frequencies can pass between.

Step-by-step solution

Define Electromagnetic Induction

Magnetic or electromagnetic induction is the process of producing an electromotive force across an electrical conductor in a shifting magnetic field. Michael Faraday discovered induction in\({\rm{1831}}\), which James Clerk Maxwell formally defined as Faraday's law of induction.

Evaluating the capacitor

The reactance of a circuit containing a capacitor is evaluated using the formula:

\({X_C}{\rm{ }} = {\rm{ }}\frac{1}{{2\pi fC}}\)

(a.) Knowing the expression for the reactance, we can solve it as:

\(C{\rm{ }} = {\rm{ }}\frac{1}{{2\pi f{X_C}}}\)

We put the numerical values and then obtain:

\(\begin{aligned} C{\rm{ }} &= {\rm{ }}\frac{1}{{2\pi {\rm{ }} \times {\rm{ }}5000{\rm{ }} \times {\rm{ }}0.01}}\\ &= {\rm{ }}3.18{\rm{ }}mF\end{aligned}\)

Therefore, the capacitance is: \(3.18{\rm{ }}mF\).

Evaluating the reactance

(b.) Knowing about the inductance, we then can now calculate what the reactance will be at another frequency. So, it is obtained as:

\(\begin{aligned} {X_C}{\rm{ }} &= {\rm{ }}\frac{1}{{2\pi {\rm{ }} \times {\rm{ }}3{\rm{ }} \times {\rm{ }}3.18{\rm{ }} \times {\rm{ }}{{10}^{ - 3}}}}\\ & = {\rm{ }}16.7{\rm{ }}\Omega \end{aligned}\)

Therefore, the reactance is: \(16.7{\rm{ }}\Omega \).

Discussing the implications

(c.) In the previous exercise, it was observed that the reactive resistance is higher for frequencies. The difference here is, that in this configuration the higher frequencies, which pass the capacitor, are shortened to the ground and do not pass between two circuits. It means that, only high frequencies can pass between them.

Therefore, it is concluded that only high frequencies can pass between.