Students studying the High School Physics Elective 3-2 (People’s Education Press edition) will encounter many new terms, such as Lenz’s Law, self-induction (inductance), inductive reactance, capacitive reactance, and so on. Regarding inductance, the Chinese Wikipedia provides the following explanation: When the current changes, an electromotive force (EMF) is generated due to electromagnetic induction to resist the change in current. Any current in a circuit produces a magnetic field, and the magnetic flux of this field in turn acts on the circuit. According to Lenz’s Law, this flux tends to resist changes in the current by means of an induced voltage (back EMF). The ratio of the change in magnetic flux to the change in current is called self-inductance, which is usually directly referred to as the inductance of the circuit.
The formula for calculating self-induction is: U = -L \frac{dI}{dt}, where U is the self-induced electromotive force, I is the current, and the negative sign indicates that the self-induced EMF opposes the original current. L is the proportionality coefficient, known as inductance. For the same coil, L is a constant with units of V \cdot t / A = \Omega \cdot t, also abbreviated as H (Henry).
Inductive Reactance
After this brief introduction, let’s get to the main topic. We know that a self-induction coil can “pass DC and block AC,” while a capacitor does the opposite, “pass AC and block DC.” We won’t dwell on the principles here; this article mainly aims to prove the calculation formulas for inductive and capacitive reactance.
First, let’s discuss inductive reactance. Since the electromotive force and current of alternating current (AC) are constantly changing and have a very small period, we usually use certain “average values” for related calculations to replace the constantly changing quantities. For example, the effective voltage (current) is the square root of the average of the squares of the voltage (current) over one period. Inductive reactance arises from this concept. Although a self-induction coil “blocks AC,” the “obstruction” at different moments within a period is generally different. Therefore, we need to calculate an “average obstruction,” which means determining what equivalent resistance this self-induction coil represents when connected to an AC circuit.
There are generally two methods to find the average value of a function: one is to average the independent variable and then substitute it into the function; the other is to directly average the function values. Inductive and capacitive reactance use the former method (after understanding this article, you might want to think about why we can only choose the former and not the latter), in other words, \bar{R} = \frac{\bar{U}}{\bar{I}} rather than \bar{R} = \overline{\left(\frac{U}{I}\right)}.
Furthermore, let the AC period be T. We have: \begin{aligned} \bar{U} &= \int_0^T \frac{|U|dt}{T} = L \int_0^T \frac{|dI|}{T} \\ \bar{I} &= \int_0^T \frac{|I| dt}{T} \end{aligned}
Next, assume the AC is sinusoidal, I = I_m \sin(2\pi f t), where the frequency is f = \frac{1}{T}. Substituting this into the above calculation, and due to the symmetry of the sine function, we only need to consider a quarter of the period. This allows us to remove the absolute value signs. That is: \begin{aligned} \bar{U} &= L \int_0^{T/4} \frac{d(I_m \sin(2\pi f t))}{T/4} = 4 I_m L f \\ \bar{I} &= \int_0^{T/4} \frac{I_m \sin(2\pi f t) dt}{T/4} = \frac{4 I_m}{2\pi} \end{aligned}
Thus, the inductive reactance X_L = \bar{R} = 4 I_m L f \div \frac{4 I_m}{2\pi} = 2\pi f L.
Capacitive Reactance
The calculation of capacitive reactance shares many similarities with inductive reactance. We know the definition of capacitance is C = \frac{Q}{U}. Since C is a constant, we can write: C = \frac{dQ}{dU} = \frac{dQ}{dt} \frac{dt}{dU}
Where \frac{dQ}{dt} is the current I, so I = C \frac{dU}{dt}. This is almost identical to the formula for self-induction! Similar to the calculation for inductive reactance, we can quickly write: \begin{aligned} \bar{I} &= \int_0^T \frac{|I|dt}{T} = C \int_0^T \frac{|dU|}{T} \\ \bar{U} &= \int_0^T \frac{|U| dt}{T} \end{aligned}
Similarly, considering a quarter period for sinusoidal AC, we have: \begin{aligned} \bar{I} &= 4 C U_m f \\ \bar{U} &= \frac{4 U_m}{2\pi} \end{aligned} The capacitive reactance X_C = \frac{\bar{U}}{\bar{I}} = \frac{1}{2\pi f C}.