Frequency Response of the RC Circuit

Thus far, the analysis of series circuits has been limited to a particular frequency. We will now examine the effect of frequency on the response of an RC series configuration such as that in [Fig. 1]. The magnitude of the source is fixed at 10 V, but the frequency range of analysis will extend from zero to 20 kHz.
Fig. 1: Determining the frequency response of a series RC circuit
Let us first determine how the impedance of the circuit $Z_T$ will vary with frequency for the specified frequency range of interest. Before getting into specifics, however, let us first develop a sense for what we should expect by noting the impedance-versus-frequency curve of each element, as drawn in [Fig. 2].
Fig. 2: The frequency response of the individual elements of a series RC circuit.
At low frequencies the reactance of the capacitor will be quite high and considerably more than the level of the resistance R, suggesting that the total impedance will be primarily capacitive in nature. At high frequencies the reactance XC will drop below the $R = 5kΩ$ level, and the network will start to shift toward one of a purely resistive nature (at 5 kΩ). The frequency at which $X_C = R$ can be determined in the following manner:
$$X_C = { 1 \over 2 \pi f_1 C} = R$$
$$\bbox[10px,border:1px solid grey]{f_1 = { 1 \over 2 \pi R C}}$$
which for the network of interest is
$$f_1 = { 1 \over 2 \pi (5kΩ)(0.01 \mu F)}=3183.1 Hz$$
For frequencies less than $f_1$, $X_C > R$, and for frequencies greater than $f_1$, $R > X_C$, as shown in Fig. 2.
Now for the details. The total impedance is determined by the following equation:
$$Z_T = R - j X_C$$
$$\bbox[10px,border:1px solid grey]{Z_T = Z_T \angle \theta_T\\ Z_T = \sqrt{ R^2 + X_C^2} \angle -tan^{-1}{X_C \over R}} \tag{1}$$
The magnitude and angle of the total impedance can now be found at any frequency of interest by simply substituting into Eq. (1). The presence of the capacitor suggests that we start from a low frequency (100 Hz) and then open the spacing until we reach the upper limit of interest (20 kHz).

f = 100 Hz

$$X_C = { 1 \over 2 \pi f C} = { 1 \over 2 \pi (100Hz)(0.01 \mu F)} = 159.16 kΩ $$
$$Z_T = \sqrt{R^2 + X^2}\\ =\sqrt{(5kΩ)^2+(159.16kΩ)^2} = 159.24 kΩ$$
$$\theta_T = -tan^{-1}{X_C \over R} = -tan^{-1}{159.16k \over 5k}\\ =-tan^{-1} 31.83 = -88.2^\circ$$
$$ Z_T = 159.24 k Ω \angle 88.2^\circ$$
which compares very closely with $Z_C = 159.16 kΩ \angle -90^\circ$ if the circuit were purely capacitive (R = 0 Ω). Our assumption that the circuit is primarily capacitive at low frequencies is therefore confirmed.

f = 1 kHz

$$X_C = { 1 \over 2 \pi f C} = { 1 \over 2 \pi (1000Hz)(0.01 \mu F)} = 15.92 kΩ $$
$$Z_T = \sqrt{R^2 + X^2}\\ =\sqrt{(5kΩ)^2+(15.92 kΩ)^2} = 16.69 kΩ$$
$$\theta_T = -tan^{-1}{X_C \over R} = -tan^{-1}{15.92k \over 5k}\\ =-tan^{-1} 3.18 = -72.54^\circ$$
$$ Z_T = 16.69 k Ω \angle -72.54^\circ$$
A noticeable drop in the magnitude has occurred, and the impedance angle has dropped almost $17^\circ$ from the purely capacitive level. Continuing:
$$f = 5 kHz: Z_T = 5.93 k Ω \angle -32.48^\circ$$ $$f = 10 kHz: Z_T = 5.25 k Ω \angle -17.66^\circ$$ $$f = 15 kHz: Z_T = 5.11 k Ω \angle -11.98^\circ$$ $$f = 20 kHz: Z_T = 5.06 k Ω \angle -9.04^\circ$$
Note how close the magnitude of $Z_T$ at $f = 20 kHz$ is to the resistance level of $5 kΩ$. In addition, note how the phase angle is approaching that associated with a pure resistive network ($0^\circ$).
A plot of $Z_T$ versus frequency in [Fig. 3] completely supports our assumption based on the curves of [Fig. 2].
Fig. 3: The magnitude of the input impedance versus frequency for the circuit of Fig. 1.

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