An RC circuit consists of a\(40.0{\rm{ }}\Omega \)resistor and a\(5.00{\rm{ }}\mu F\)capacitor.

(a) Find its impedance at\(60.0{\rm{ }}Hz\)and\(10.0{\rm{ }}kHz\).

(b) Compare these values of\(Z\)with those found in Example\(23.12\), in which there was also an inductor.

A circuit is a closed path via which electricity can flow from one location to another. It could include transistors, resistors, and capacitors, among other electrical components.

a)

Let us consider the given information,

Formula used:\({X_C} = \frac{1}{{2\pi fC}}\)

Here\({X_C}\)is capacitive reactance,\(f\)is the frequency and\(L\)is the capacitance.

For each frequency, impedance is given by:

\(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \)

At high frequency, we have,\(f = 60\;{\rm{Hz}},\,\,C = 5.0 \times {10^{ - 6}}\;{\rm{F}}\)and\({X_L} = 0\)(there is no inductor)

Substituting the given values in above equation, we get

\(\begin{aligned} {X_C} &= \frac{1}{{2 \times 3.14 \times \left( {60} \right)\left( {5 \times {{10}^{ - 6}}} \right)}}\\ &= 530.52\,\Omega \end{aligned}\)

Impedance,

\(\begin{aligned} Z &= \sqrt {{{(40)}^2} + {{(530.52)}^2}} \\Z &= 532\,\Omega \end{aligned}\)

At low frequency, substituting the given values in above equation, we get

\(\begin{aligned} X_C^\prime &= \frac{1}{{2 \times 3.14 \times \left( {1 \times {{10}^4}} \right)\left( {5 \times {{10}^{ - 6}}} \right)}}\\ &= 3.18\,\Omega \end{aligned}\)

Impedance,

\(\begin{aligned} {Z^\prime } &= \sqrt {{{(40)}^2} + {{(3.18)}^2}} \\{Z^\prime } &= 40.1\,\,\Omega \end{aligned}\)

Therefore, impedance at low and high frequencies are \(188.5\Omega \)and \(193\Omega \)

b)

Apply the formula.

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

Here v is capacitive reactance, \({\rm{f}}\)is the frequency and \({\rm{L}}\) is the capacitance.

For each frequency, impedance is given by:

\(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \)

Calculation:

As we can see, at\(60\;{\rm{Hz}}\,\,Z\)is about the same with or without the inductor because the capacitor dominates at low frequency. At\(10.0{\rm{kHz}}\,\,Z\)is much less without the inductor because the inductor dominates at high frequencies.

Therefore, at \(10.0{\rm{kHz}}\,\,{\rm{Z}}\)is much less without the inductor because the inductor dominates at high frequencies