Question

The capacitor in Figure \(23.55\) (a) is designed to filter low-frequency signals, impeding their transmission between circuits. (a) What capacitance is needed to produce a \(100{\rm{ }}k\Omega \) reactance at a frequency of \(120{\rm{ }}Hz\)? (b) What would its reactance be at \(1.00{\rm{ }}mHz\)? (c) Discuss the implications of your answers to (a) and (b).

Short answer

(a.) The capacitance is obtained as: \(13.3{\rm{ }}nF\).

(b.) The reactance is obtained as: \(12{\rm{ }}\Omega \).

(c.) The capacitor tends to act as a high-pass filter.

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:

\(\begin{aligned} C{\rm{ }} &= {\rm{ }}\frac{1}{{2\pi {\rm{ }} \times {\rm{ }}120{\rm{ }} \times {\rm{ }}1{\rm{ }} \times {\rm{ }}{{10}^5}}}\\ &= {\rm{ }}13.3{\rm{ }}nF\end{aligned}\)

Therefore, the capacitance is: \(13.3{\rm{ }}nF\).

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:

\(\begin{aligned} {X_C}{\rm{ }} &= {\rm{ }}\frac{1}{{2\pi {\rm{ }} \times {\rm{ }}1{\rm{ }} \times {\rm{ }}{{10}^6}{\rm{ }} \times {\rm{ }}1.33{\rm{ }} \times {\rm{ }}{{10}^{ - 8}}}}\\ &= {\rm{ }}12{\rm{ }}\Omega \end{aligned}\)

Therefore, the reactance is: \(12{\rm{ }}\Omega \).

Discussing the implications

(c.) We observe in this exercise that the reactive resistance is very much higher for lower frequencies. Which means, higher frequencies can pass while lower ones will do the same in much lower intensities. That concludes, this capacitor acts as a high-pass filter. Low frequencies will not pass from the first circuit to the second circuit, on the other hand the higher ones will.

Therefore, it is concluded that capacitor tends to act as a high-pass filter.