A resistor is a 2-terminal device that satisfies Ohm's law:$$\begin{equation}V(t)=I(t)R\end{equation}$$ where $R$ is the resistance and $V(t)$ is the voltage from the terminal where current $I(t)$ is assumed to enter the resistor to the terminal where the current is assumed to exit the resistor. In the case of DC circuits (i.e. when the voltage and current do not depend on time) the above equation is usually written as $V=IR$, where $V$ and $I$ are the DC values of the voltage and current. In the case of AC circuits (in frequency domain) the Ohm's law can be written in the same way, but $V$ and $I$ are the complex values of the voltage and current.

• If the resistance is infinite, we have an open circuit. In this case the current will be zero and the resistor can be removed without affecting the rest of the circuit.
• If the resistance is zero, we have a short circuit. In this case the voltage will be zero and the resistor can be replaced with an ideal wire without affecting the rest of the circuit.

Why do resistors warm up? When current flows through a resistor, the moving charge carriers (usually electrons) collide with the atoms of the resistive material. These collisions impede the flow of electrons and convert part of the electrical energy into random thermal motion of the atoms — i.e., heat and possibly into light. This process in which the electrical energy is converted into heat is called Joule heating (or resistive heating).

A capacitor is a 2-terminal device in which the current and voltage satisfy the following equation $$\begin{equation}I(t)=C\dfrac{dV(t)}{dt}\end{equation}$$ where $C$ is the capacitance (see figure for notations). In frequency domain (AC analysis) this equation becomes $$\begin{equation}I=\frac{V}{j{\omega}C}\end{equation}$$ where $I$ and $V$ are the complex values of the current and voltage and $\omega$ is the angular frequency of the AC signal (the above equation is equivalent to Ohm's law).

Can the voltage across a capacitor change instantaneously? The answer is no, because doing so would cause the voltage across the inductor to become infinite (see the equation above). As a result, the current through an inductor cannot change instantaneously. In practice, one should always avoid sudden changes in the current through an inductor—for example, abruptly disconnecting the inductor from the circuit. In such a case, forcing the current to drop instantaneously from its initial value to zero can produce a very large voltage spike, potentially damaging the inductor or other circuit components. Instead, the energy stored in the inductor’s magnetic field will try to keep the current flowing, often discharging through a spark, arc, or protection device if no other path is provided.

Capacitors can be divided into polarized and non-polarized types:

• Polarized capacitors are designed to be connected with a specific polarity (a positive and a negative lead). Applying reverse voltage can damage them or cause them to fail. The most common polarized types are:
  1. Electrolytic capacitors (aluminum electrolytic, tantalum, niobium): by far the most widely used; has a capacitance of the order of microfarads to milifarads.
  2. Supercapacitors or ultracapacitors: an electrochemical device with an extremely high capacitance (of the order of farads or thousands of farads) that is based on an electric double layer.
• Non-polarized capacitors (film, ceramic, mica, paper, etc.) can be connected in either direction and do not have polarity markings. These are used where AC or changing polarity signals are present.

An inductor is a 2-terminal device in which the current and voltage satisfy the following equation $$\begin{equation}V(t)=L\dfrac{I(t)}{dt}\end{equation}$$ where $L$ is the inductance (see figure for notations). In frequency domain (AC analysis) this equation becomes $$\begin{equation}V=j{\omega}L I\end{equation}$$ where $I$ and $V$ are the complex values of the current and voltage and $\omega$ is the angular frequency of the AC signal (the above equation is equivalent to Ohm's law).

Can the current through an inductor change instantaneously? The answer is no, because doing so the voltage across the inductor would become infinite (see the equation above). As a result, the current through an inductor cannot change instantaneously. In practice, one should always avoid sudden changes in the current through an inductor—for example, suddently disconnecting the inductor from the circuit. In such a case, the current would decrease instantaneously from the initial value to 0, which could damage the inductor..

Can the current through an inductor change instantaneously? The answer is no, because doing so the voltage across the inductor would become infinite (see the equation above). As a result, the current through an inductor cannot change instantaneously. In practice, one should always avoid sudden changes in the current through an inductor—for example, suddently disconnecting the inductor from the circuit. In such a case, the current would decrease instantaneously from the initial value to 0, which could damage the inductor..

An independent voltage source is a 2-terminal device in which the voltage from the positive to the negative terminals is equal to $V$. The current flowing through a voltage source is generally unknown until the circuit is solved analytically (or the current is measured experimentally).

If a voltage source is short-circuited, the circuit presents almost zero resistance to the source. As a result, the current rises dramatically, often far exceeding the source’s rated output. In general, this should never be done, as the excessive current can damage the source unless it is equipped with an appropriate protection mechanism.

An independent current source is a 2-terminal device in which the current flowing in the direction of the arrow is equal to $I$. The voltage accross a current source is generally unknown until the circuit is solved analytically (or the voltage is measured experimentally).

Note that a current source adjusts the voltage across its terminals to maintain the current at its specified value. But what happens if a current source is disconnected from the circuit so that its terminals are left open? In practice, one should never do this, because with no path for current flow the source may develop an excessively high voltage and could be damaged unless it is equipped with a protection mechanism.