# Frequency Response of the Basic Elements

The analysis of the previous section was limited to a particular applied frequency. What is the effect of varying the frequency on the level of opposition offered by a resistive, inductive, or capacitive element? We are aware from the last section that the inductive reactance increases with frequency while the capacitive reactance decreases. However, what is the pattern to this increase or decrease in opposition? Does it continue indefinitely on the same path? Since applied signals may have frequencies extending from a few hertz to megahertz, it is important to be aware of the effect of frequency on the opposition level.

### R

Thus far we have assumed that the resistance of a resistor is independent of the applied frequency. However, in the real world each resistive element has stray capacitance levels and lead inductance that are sensitive to the applied frequency. However, the capacitive and inductive levels involved are usually so small that their real effect is not noticed until the megahertz range. The resistance-versus-frequency curves for a number of carbon composition resistors are provided in Fig. 1 Note that the lower resistance levels seem to be less affected by the frequency level. The 100Ω resistor is essentially stable up to about 300 MHz, whereas the 100kΩ resistor starts its radical decline at about 15 MHz.
Fig. 1: Typical resistance-versus-frequency curves for carbon compound resistors.
Frequency, therefore, does have impact on the resistance of an element, but for our current frequency range of interest, we will assume the resistance-versus-frequency plot of Fig. 2 (like Fig. 1 up to 15 MHz), which essentially specifies that the resistance level of a resistor is independent of frequency.
Fig. 2: R versus f for the range of interest.

### L

For inductors, the equation $$X_L = wL = 2\pi fL = 2\pi Lf$$ is directly related to the straight-line equation $$y = mx+b =( 2\pi L )f+0$$ with a slope (m) of $2\pi L$ and a y-intercept (b) of zero. $X_L$ is the y variable and f is the x variable, as shown in Fig. 3.
Fig. 3: XL versus frequency.
The larger the inductance, the greater the slope ($m = 2\pi L$) for the same frequency range, as shown in Fig. 3. Keep in mind, as reemphasized by Fig. 3, that the opposition of an inductor at very low frequencies approaches that of a short circuit, while at high frequencies the reactance approaches that of an open circuit.

### C

For the capacitor, the reactance equation $$X_C = {1 \over 2\pi f C}$$ can be written $$X_Cf = {1 \over 2\pi C}$$ which matches the basic format of a hyperbola, $$yx = k$$ with $y = X_C$, $x = f$, and the constant $k = 1/(2\pi C)$.
At $f = 0 Hz$, the reactance of the capacitor is so large, as shown in Fig. 4, that it can be replaced by an open-circuit equivalent.
Fig. 4: XC versus frequency.
As the frequency increases, the reactance decreases, until eventually a short circuit equivalent would be appropriate. Note that an increase in capacitance causes the reactance to drop off more rapidly with frequency.
In summary, therefore, as the applied frequency increases, the resistance of a resistor remains constant, the reactance of an inductor increases linearly, and the reactance of a capacitor decreases nonlinearly.
Example 1: At what frequency will the reactance of a 200mH inductor match the resistance level of a 5-kΩ resistor?
Solution: The resistance remains constant at 5 kΩ for the frequency range of the inductor. Therefore, $$R = 5000 Ω = X_L = 2\pi fL\\ = 2\pi (200 \times 10^{-3} H)f = 1.257f$$ and $$f = {5000 \over 1.257}= 3.98kHz$$
Example 2: At what frequency will an inductor of 5 mH have the same reactance as a capacitor of $0.1 \mu F$?
Solution: $$\begin{split} X_L &= X_C \\ 2\pi fL &= { 1\over 2 \pi fC}\\ f^2 &= {1 \over 4 \pi^2 LC}\\ f &= {1 \over 2 \pi \sqrt{LC}}\\ f &= {1 \over 2 \pi \sqrt{5mH \times 0.1\mu F}}\\ f &= 7.12kHz\\ \end{split}$$