Linear Circuit Analysis is a core course in electrical engineering that most students need to take in the first year of their program. In fact, the course covers such fundamental topics that, most often, students from other engineering areas such as computer, industrial, power, or manufacturing engineering also need to take it as an introduction to electrical engineering.

This webbook is designed to be used by students to learn and solve problems on the CircuitsU website. The webbook is slightly different from other books or materials in linear circuit analysis that you can find online. First, the webbook is written in a concise manner to make students spend a minimum amount of time learning the different topics on linear circuits and be able to spend more time practicing problems on their own. For this reason, just by browsing the pages of the webbook might not be sufficient to master all the topics on linear circuits (like a regular textbook is supposed to do) and students might need to first look at the Sample Solved Problems sections that appears at the of most section, then, practice problems from the Practice Problems sections.

Second, the problems and examples that appear at the end of each section are generated randomly when the page is loaded, so students can look over and practice as many problems as they want. Since the webbook was written specifically for this website, you need to be logged in order to take advantage of the full set of examples and problems that appear in the webbook.

This webbook and the problems that CircuitsU can generate automatically, cover most topics in the Linear Circuit Analysis course taught in undergraduate programs in the U.S. and is organized relatively close to the existing textbooks on linear circuits. Therefore, the instructors can use CircuitsU to teach and assign homework problems in linear circuits.

Finally, please keep in mind that this webbook is still under construction. You might find typos or unfinished sections.

Below is the list of chapters, for which CircuitsU can generate random problems. The instructor can re-organize the content of the chapters and topics covered in each chapter, and change the type, difficulty, and number of homework problems that the site will generate in each homework assignment. The instructor can also build homework assignments covering problems from different chapters (for instance, to design a midterm or a final exam).

Chapter 1 is an introduction to electric components and circuit diagrams and is the homework associated with it is usually assigned in the first week of classes for students to get used with the website. Chapters 2-9 focus on DC circuits; chapters 10-21 focus on AC circuits; chapters 22-27 focus on Laplace transforms and first and second-order transient circuits. These homework assignments cover most of the topics that are usually covered in a college-level, introductory course on linear circuit analysis. For instance, if the Linear Circuit Analysis course is taught as a:

• year-long course sequence: most topics could be covered throughout the course
• two-semester course sequence: chapters 1-16 could be offered during the first semester and chapters 17-26 are offered during the second semester
• three-quarter course sequence: chapters 1-9 could be offered during the first quarter, chapters 10-17 are offered during the second quarter, and chapters 18-26 are offered in the third quarter

• General problems, in which students need to answer simple questions such as identifing different components in a circuit, or series and parallel connections.
• Numerical problems, (ℕ), in which the known variables in the problem are given numerically and students need to compute the shought variable or variables numerically.
• Analytical problems, (Ⓐ), in which the known variables are given symbolically and the students need to compute the analytically.
• Design problems, (ⅅ), in which students need to comptute specific elements in a circuit in order for some conditions to be satisfied (e.g. what are the values of specific resistors such that the gain of an amplifier to be equal to 10).

Below are the recommended assignments (HW1-HW31) that are generated by default when the instructor creates a new Linear Circuit Analysis course and choses Selected HW assignments by chapter (32 assignments).

Below are the recommended assignments (HW1-HW17) that are generated by default when the instructor creates a new Linear Circuit Analysis course and choses Semester 1 in a 2-semester course. The table shows the proposed due dates for each HW set but each instructor can set his or her own time schedule. The instructor can also re-organize the content and change the type, difficulty, and number of homework problems in each set.

Below are the recommended assignments (HW1-HW11) that are generated by default when the instructor creates a new Linear Circuit Analysis course and choses Semester 2 in a 2-semester course.

^aThe week when the topic is covered.

^bHomeworks that have the same assignment number have the same deadlines. For instance, Assignment number 1 containts HW1, HW2, and HW3, which are all due at the end of the second week of classes.

Below are the recommended assignments (HW1-HW10) that are generated by default when the instructor creates a new Linear Circuit Analysis course and choses Quarter 2 in a 3-quarter course.

Below are the recommended assignments (HW1-HW10) that are generated by default when the instructor creates a new Linear Circuit Analysis course and choses Quarter 3 in a 3-quarter course.

This webbook and CircuitsU use the following notations:

^*A better notation, which is often used in textbooks, is to use bold letters for complex variables ($\boldsymbol{I}$, $\boldsymbol{V}$, $\boldsymbol{P_d}$, and $\boldsymbol{P_g}$) however, due to limitations in SVG graphic fonts we are representing them with regular fonts on this website.

All variables are assumed to be expressed in the International System of Units in CircuitsU. Therefore, to simplify notations, CircuitsU will usually not write the units in mathematical expressions unless it is a final answer.

Units are written with gray characters in CircuitsU. For instance: $V_1 = 2.4 \: {\textcolor{gray}V}$, $R = \frac{2.4 \: {\textcolor{gray}V}}{1.2 \: {\textcolor{gray}A}} =2 \: {\textcolor{gray}Ω}$.

Table 2.1 shows a list of linear components (also called devices or elements) used in linear circuit analysis. All the components (and, implicitly, their current-voltage characteristics) are assumed to be ideal.

A loop (or mesh is any closed path through the circuit in which nodes appear at most once. Meshes are defined in planar circuits, which are circuits that can be drawn on a plane surface with no wires crossing each other. Loops are slightly more general and can be applied to any circuit, planar or not. Since all the electric circuits that appear on this website are planar we will use mesh and loop interchangeably.

The outside mesh of a circuit is the largest mesh in the circuit; a minor mesh (also called essential mesh) is a mesh that does not contain any other meshes.

A circuit can have multiple loops (or meshes). For instance, the circuit shown in Fig. 2.2 has 3 minor meshes labeled $i_1$, $i_2$, and $i_3$ and the outside mesh. Since, we are almost always interested only in the minor meshes and the outside mesh, we will simply say that that the circuit below contains 3 meshes (or 3 loops) plus the outside mesh (or outside loop).

Loops and meshes are made of branches.

Any circuit should contain at least 1 mesh. When the circuit contains only 1 mesh (excluding the outside mesh), it is called single loop circuit (see Fig. 2.3).

Mesh analysis is a method based on Kirchhoff’s Voltage Law (KVL) that can be used to analyze planar electric circuits. Loop analysis is slightly more general and can be applied to non-planar circuits. Mesh analysis is usually easier to use than loop analysis because the circuit is planar. However, notice again that, since the electric circuits on this site are planar, we will use mesh analysis and loop analysis interchangeably.

A node is a collection of wires that are connected to each other. A circuit can have multiple nodes. For instance, the circuit shown in Fig. 2.4 has 5 nodes labeled $v_1$, $v_2$, $v_3$, $v_4$, and $v_5$.

Sometimes, one of the nodes is called the ground node. For instance, the circuits showed in Fig. 2.5 and Fig. 2.6 contain 4 nodes labeled $v_1$, $v_2$, $v_3$, and $v_4$, and the ground node (5 nodes in total). Notice that the two circuits have the same topology and will have the same solution.

Any circuit should contain at least 2 nodes (see Fig. 2.7). When the circuit contains only 2 nodes, it is called a single node-pair circuit.

Nodal analysis is a method based on Kirchhoff’s Current Law (KCL) that can be used to analyze electric circuits.

Two or more components (each with two terminals) are connected in series when they are part of the same two minor meshes, or, to one minor mesh and the outside mesh. For instance, in Fig. 2.8:

• Voltage source $V_1$ and resistor $R_1$ are connected in series because they both belong to minor mesh $i_1$ and the outside mesh.
• Current source $I_1$ and capacitor $C_1$ are connected in series because they both belong to minor mesh $i_3$ and the outside mesh.
• However, resistor $R_3$ and inductor $L_1$ are not connected in series because they belong to different pair of minor meshes (resistor $R_3$ belongs to $i_1$ and $i_2$, while inductor $L_1$ belongs to the outside mesh and $i_2$.

In Fig. 2.9:

• Voltage source $V_1$ and inductor $L_1$ are connected in series because they both belong to minor mesh $i_2$ and the outside mesh.
• Capacitor $C_1$, current source $i_1$ and capacitor $C_2$ are connected in series because they all belong to minor mesh $i_3$ and the outside mesh.
• Resistor $R_2$ and inductor $L_2$ are connected in series because they both belong to minor meshes $i_2$ and $i_3$.

If one component is connected in series with a second component, and the second component is connected in series with the third component, then, the first component is also connected in series with the third component.

Two or more components (each having two terminals) are connected in parallel when they are connected between the same two nodes. For instance, looking at the circuit in Fig. 2.9:

• Capacitor $C_3$ and resistor $R_1$ are connected in parallel because they are both connected between nodes $v_2$ and $v_3$.
• However, inductors $L_1$ and $L_2$ are not connected in parallel because they are not connected between the same two nodes (inductor $L_1$ is connected between $v_3$ and $v_6$, while inductor $L_2$ is connected between $v_4$ and $v_6$.

If one component is connected in parallel with a second component, and the second component is connected in parallel with the third component, then, the first component is also connected in parallel with the third component.

If two or more resistors $R_1$, $R_2$, ... $R_n$ are connected in series they are equivalent with a resistor with $$R_{eff}=R_1+R_2+...+R_n$$ That means we can replace one of the $n$ resistors with an equivalent resistor with resistance $R_{eff}$ and replace the other resistors with short-circuits (wires).

For instance, considering the circuit in Fig. 2.10, resistors $R_1$ and $R_2$ are connected in series. Therefore, we can keep one the resistors, say $R_1$, replace its value with $R_{eff}=R_1+R_2$, and replace resistor $R_2$ with a wire.

If two or more resistors $R_1$, $R_2$, ... $R_n$ are connected in parallel they are equivalent with a resistor with $$R_{eff}=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}}$$ That means we can replace one of the $n$ resistors with an equivalent resistor with resistance $R_{eff}$ and remove all the other resistors. When we have only two resistors connected in parallel, we can replace them with a single resistors with resistance $$R_{eff}=\frac{R_1 R_2}{R_1+R_2}$$

For instance, considering the circuit in Fig. 2.11, resistors $R_3$, $R_4$, and $R_5$ are connected in parallel. Therefore, we can keep one the resistors, say $R_3$, replace its value with $R_{eff}=\frac{1}{\frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_5}}$, and remove resistor $R_4$ from the circuit. Similarly, we could keep $R_4$ and remove $R_3$ and $R_5$, or keep $R_5$ and remove $R_3$ and $R_4$.

If two or more inductors $L_1$, $L_2$, ... $L_n$ are connected in series they are equivalent with a inductor with $$L_{eff}=L_1+L_2+...+L_n$$ That means we can replace one of the $n$ inductors with an equivalent inductor with inductance $L_{eff}$ and replace the other inductors with short-circuits (wires).

For instance, considering the circuit in Fig. 2.12, inductors $L_1$ and $L_2$ are connected in series. Therefore, we can keep one the inductors, say $L_1$, replace its value with $L_{eff}=L_1+L_2$, and replace inductor $L_2$ with a wire.

If two or more inductors $L_1$, $L_2$, ... $L_n$ are connected in parallel they are equivalent with a inductor with $$L_{eff}=\frac{1}{\frac{1}{L_1}+\frac{1}{L_2}+...+\frac{1}{L_n}}$$ That means we can replace one of the $n$ inductors with an equivalent inductor with inductance $L_{eff}$ and remove all the other inductors. When we have only two inductors connected in parallel, we can replace them with a single inductors with inductance $$L_{eff}=\frac{L_1 L_2}{L_1+L_2}$$

For instance, considering the circuit in Fig. 2.13, inductors $L_3$, $L_4$, and $L_5$ are connected in parallel. Therefore, we can keep one the inductors, say $L_3$, replace its value with $L_{eff}=\frac{1}{\frac{1}{L_3}+\frac{1}{L_4}+\frac{1}{L_5}}$, and remove inductor $L_4$ from the circuit. Similarly, we could keep $L_4$ and remove $L_3$ and $L_5$, or keep $L_5$ and remove $L_3$ and $L_4$.

If two or more capacitors $C_1$, $C_2$, ... $C_n$ are connected in series they are equivalent with a capacitor with $$C_{eff}=\frac{1}{\frac{1}{C_1}+\frac{1}{C_2}+...+\frac{1}{C_n}}$$ That means we can replace one of the $n$ capacitors with an equivalent capacitor with capacitance $C_{eff}$ and replace the other capacitors with short-circuits (wires). When we have only two capacitors connected in parallel, we can replace them with a single capacitors with capacitance $$C_{eff}=\frac{C_1 C_2}{C_1+C_2}$$

For instance, considering the circuit in Fig. 2.14, capacitors $C_1$ and $C_2$ are connected in series. Therefore, we can keep one the capacitors, say $C_1$, replace its value with $C_{eff}=\frac{C_1 C_2}{C_1+C_2}$, and replace capacitor $C_2$ with a wire.

If two or more capacitors $C_1$, $C_2$, ... $C_n$ are connected in parallel they are equivalent with a capacitor with $$C_{eff}=C_1+C_2+...+C_n$$ That means we can replace one of the $n$ capacitors with an equivalent capacitor with capacitance $C_{eff}$ and remove all the other capacitors.

For instance, considering the circuit in Fig. 2.15, capacitors $C_3$, $C_4$, and $C_5$ are connected in parallel. Therefore, we can keep one the capacitors, say $C_3$, replace its value with $C_{eff}=C_3+C_4+C_5$, and remove capacitors $C_4$ and $C_5$ from the circuit. Similarly, we could keep $C_4$ and remove $C_3$ and $C_5$, or keep $C_5$ and remove $C_3$ and $C_4$.

If two or more voltage sources $V_1$, $V_2$, ... $V_n$ are connected in series they can be replaced with a single voltage source with $$V_{eff}=\pm V_1 \pm V_2\pm...\pm V_n$$ where the terms in the right hand side are taken with $+$ sign if the corresponding voltage source $V_i$ is oriented in the same direction with $V_{eff}$ and with $-$ sign if $V_i$ is oriented in opposite direction with $V_{eff}$. Since the voltage sources are all connected in series, when we replace $V_i$ with $V_{eff}$, we need to replace the other voltage sources with short-circuits (wires).

For instance, considering the circuit in Fig. 2.16, voltage sources $V_1$ and $V_2$ are connected in series. Therefore, we can keep one the voltage sources, say $V_1$, replace its value with $V_{eff}=V_1-V_2$, and replace voltage source $V_2$ with a wire.

Voltage sources should never be combined in parallel.

Current sources should never be combined in series.

If two or more current sources $I_1$, $I_2$, ... $I_n$ are connected in parallel they can be replaced with a single current source with $$I_{eff}=\pm I_1 \pm I_2\pm...\pm I_n$$ where the terms in the right hand side are taken with $+$ sign if the corresponding current source $I_i$ is oriented in the same direction with $I_{eff}$ and with $-$ sign if $I_i$ is oriented in opposite direction with $I_{eff}$. Since all the current sources are connecte in parallel, when we replace $I_i$ with $I_{eff}$, we need to remove the other current sources.

For instance, considering the circuit in Fig. 2.17, current sources $I_1$, $I_2$, and $I_3$ are connected in parallel. Therefore, we can keep one the current sources, say $I_1$, replace its value with $I_{eff}=I_1-I_2+I_3$, and remove current sources $I_2$ and $I_3$ from the circuit.

The power dissipated by a two terminal component is equal to

where voltage $V(t)$ and current $I(t)$ are shown in Fig. 2.19. Notice that we used the same sign convention for $V(t)$ and $I(t)$ as in Ohm's law. In general, the power dissipated by a component is a time-dependent quantity and depends on the instanteneous values of the voltage and current.

In the case of resistors, the power dissipated becomes $$\begin{equation}P_d(t)=R~I(t)^2=\frac{V(t)^2}{R}\end{equation}$$where we used Ohm's law. Notice that, since we square both the current and voltage, the power dissipated by a resistors is non-negative (assuming that the resistance is positive). At steady-state (in the case of DC circuits), the power dissipated does not depend on time and $P_d=RI^2=\frac{V^2}{R}$.

The power dissipated by a component is measured in watts (W).

The power generated by a two terminal component is negative the power dissipated by that component

where voltage $V(t)$ and current $I(t)$ are shown in Fig. 2.18.

The power generated by a component is measured in watts (W).

Consider an arbitrary lumped network that has $b$ branches and $n$ nodes. Suppose that to each branch we assign arbitrarily a branch potential difference $V_k$ and a branch current $I_k$ for $k=1,...,b$ and suppose that they are measured with respect to arbitrarily picked associated reference directions. Tellegen's theorem states that $$\begin{equation}\sum_{k=1}^{b}{V_k I_k}=0\end{equation}$$

Tellegen's theorem shows that, if KVL and KCL are satisfied, the algebraic sum of the powers dissipated by an isolated electric circuit is equal to 0. In other words, the total power dissipated in a circuit is equal to the total power generated in the circuit.

The energy dissipated by a two terminal component from moment $t_0$ until $t$ is equal to $$\begin{equation}E_d=\int_{t_0}^{t} \, V(t)I(t) \, dt\end{equation}$$

where voltage $V$ and current $I$ are shown in Fig. 2.20. Notice that we used the same sign convention for $V$ and $I$ as in Ohm's law.

At steady-state (in the case of DC circuits), the energy dissipated dobecomes $E_d=IV(t-t_0)$. Notice that the energy dissipated increases linearly as times passes.

The energy dissipated by a component is measured in joules (J).

The energy generated by a two terminal component is negative the energy dissipated by that component $$\begin{equation}E_g=-E_d\end{equation}$$ and is measured in joules (J).

Ohm’s law states that the current through a resistor is directly proportional to the voltage across the resistor. Denoting the constant of proportionality with $R$ (the resistance), one can express Ohm's law mathematically as $$\begin{equation}V=RI\end{equation}$$ or $$\begin{equation}I=\frac{V}{R}\end{equation}$$ where voltage $V$ and current $I$ are shown in Fig. 3.20. It is important to note that the value of the current that we obtain using Ohm's law depends on how we define voltage $V$. As shown in the figure, the current obtained using Ohm's law is assumed to flow from the $+$ to the $-$ nodes (however, depending on the value of $V$ this current can be either positive or negative).

If we introduce the conductance as the inverse of the resistance $C=1/R$, Ohm's law becomes $V=I/C$ or $I=CV$. The unit for resistance is ohm ($Ω$); the unit for conductance is siemens ($S$), also known as mho.

Voltage division is a technique that can be used to compute the voltage across each resistor of a series combination of resistors when the voltage across all the resistors is known. For instance, if we have $n$ resistors connected in series and the total voltage across them is $V$, the voltage across resistor $R_i$ is equal to $$\begin{equation}V_i=V\frac{R_i}{R_1+R_2+...+R_n}\end{equation}$$

In the case of only two resistors, the previous equation gives $$\begin{equation}V_1=V\frac{R_1}{R_1+R_2}\end{equation}$$ $$\begin{equation}V_2=V\frac{R_2}{R_1+R_2}\end{equation}$$

For instance, consider the circuit shown in Fig. 3.26.

Applying voltage division we obtain $$V_{1}=20\ V \times\frac{2}{2+5+3}=4\ V$$ $$V_{2}=20\ V \times\frac{5}{2+5+3}=10\ V$$ $$V_{3}=-20\ V \times\frac{3}{2+5+3}=-6\ V$$ Notice that, in the last equation, we took $V_3$ with negative sign because voltage $V_3$ is defined in opposite direction than the voltage induced by the $20\ V$ voltage source.

Current division is a technique that can be used to compute the current going through each resistor of a parallel combination of resistors when the current going through all the resistors is known. In general, if we have $n$ resistors connected in parallel and the total current going through them is $I$, the current going through resistor $R_i$ is equal to $$\begin{equation}I_i=I\frac{\frac{1}{R_i}}{\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}}\end{equation}$$

In the case of only two resistors, the previous equation gives $$\begin{equation}I_1=I\frac{R_2}{R_1+R_2}\end{equation}$$ $$\begin{equation}I_2=I\frac{R_1}{R_1+R_2}\end{equation}$$

For instance, consider the circuit shown in Fig. 3.27.

Applying current division we obtain $$I_{1}=6\ A \times\frac{\frac{1}{8}}{\frac{1}{4}+\frac{1}{8}}=4\ A$$ $$I_{2}=-6\ A \times\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{8}}=-2\ A$$ Notice that, in the last equation, we took $I_2$ with negative sign because the current flows in opposite direction than the current induced by the $6\ A$ current source.

Superposition is a method that can be used to analyze linear electric circuits. The method is based on the fact that the values of the potentials and currents in a circuit containing multiple independent sources can be calculated by adding algebraically the potentials and currents obtained by activating one source at a time.

Assume we have a circuit that contains at least two independent sources.

Step 1. Keep only one independent sources and deactivate all the other independent sources. Notice that if you circuit contains dependent sources you need to keep those in the circuit (in a way, they are treated like resistors).

Step 2. Calculate the potentials and currents in the circuit.

Step 3. Repeat the last two steps for each source in the circuit.

Step 4. Superimpose the values of potentials and currents obtained at Step 2; pay special attention to the direction of the voltage drops and current flows.

• To deactivate a current source we remove it from the circuit (replace it with open-circuit or break); to deactivate a voltage source we replace it with a short-circuit (or wire).
• Unlike mesh and nodal analysis that can be applied to both nonlinear and linear circuits, superposition can be applied only to linear circuits. Therefore, we cannot use superposition to compute the potential and currents in a circuit containing diodes, transistors, or other nonlinear elements.
• The superposition method can be applied to DC, AC and time-dependent circuits (as long as the circuit is linear).
• The superposition method cannot be used to add power. If we need to compute the power generated or dissipated by a component, we need to compute the voltage across and current flowing through the component first and, then, the power.

Source transformation is the process of simplifying a linear circuit, especially with mixed sources, by transforming independent voltage sources into independent current sources using Norton's theorem, and vice versa using Thévenin's theorem . The process is usually combined with series and parallel simplifications of resistor (or impedance in the case of AC circuits), current sources and voltage sources.

When transforming a circuit from a Norton equivalent circuit to a Thévenin equivalent circuit the value of the resistance remains the same, while the voltage of the voltage source is $$\begin{equation}V=R I\end{equation}$$

When transforming a circuit from a Thévenin equivalent circuit to a Norton equivalent circuit the value of the resistance remains the same, while the current of the current source is $$\begin{equation}I=\frac{V}{R}\end{equation}$$

Have you ever wondered how an ohmmeter works (an ohmmeter is an instrument that measures the resistance or impedance of a circuit)? In order to measure the resistance of a network the ohmmeter applies a small voltage at the terminals of the network and measure the current going through the terminals. Then, it uses Ohm's law to compute the resistance of the circuit. This principle stays at the basis of the $V_{test}/I_{test}$ method that is often used in electronics to measure or compute the resistance of a network. Depending on the type of the source that is used to energize the network, we distinguish two methods.

When using the test voltage source method we apply a test voltage (in theoretial calculations often taken equal to $1 \: {\textcolor{gray}V}$) and compute the value of the current injected by this source. Then, the resitance of the network is comptued using

When using the test current source method we apply a test current (in theoretial calculations often taken equal to $1 \: {\textcolor{gray}A}$) and compute the value of the voltage at the terminals of the network. Then, the resitance of the network is comptued using the same equation as above.

1. Whenever you use the $V_{test}/I_{test}$ method to determine the resistance of a circuit, you need to make sure your circuit does not have any independent sources that could affect your measurements (in practice they can demage your ohmmeter).
2. The $V_{test}/I_{test}$ method is often used to determine the Norton and Thévenin resistance (or impedance in the case of AC circuits) of a network.

Norton's theorem states that any two-terminal linear circuit containing only voltage sources, current sources and resistors (or impedances in the case of AC circuits) can be replaced by an equivalent combination of a current source $I_{N}$ in parallel with a resistor $R_{N}$ (or impedance $Z_{N}$ in the case of AC circuits).

Norton's theorem and its dual, Thévenin's theorem, are widely used to simplify circuit analysis and study a circuit's initial-condition and steady-state response. Norton's theorem may in some cases be more convenient to use than Kirchhoff's circuit laws when analyzing linear circuits.

Calculating the Norton equivalent circuit, or simply the Norton equivalent of a circuit with two terminals means computing the Norton current and Norton resistance of the circuit.

The Norton current can be computed using any of the following methods:

• $I_{N} = I_{s.c}$ - the Norton current is equal to the short-circuit current at the output terminals of the original circuit. To compute $I_{s.c.}$ one can use any of the methods of linear circuit analysis such as nodal analysis, mesh analysis, or current and voltage division.
• If the Thévenin voltage $V_{Th}$ and Thévenin resistance $R_{Th}$ are known and $R_{Th} \neq 0$, one can calculate the Norton current using $I_{N} = V_{Th}/R_{Th}$ .

The Norton resistance can be computed using any of the following methods:

• $R_{N}$ can be computed by deactivating all independent sources and using the test voltage or test current methods using Ohm's law, $R_{N}=\frac{V_{test}}{I_{test}}$. To use the test voltage method, we connect a test voltage source at the output terminals (usually, but not necessarily, taken as $V_{test}=1~V$) and compute current $I_{test}$. To use the test current method, we connect a test current source at the output terminals (usually, but not necessarily, taken as $I_{test}=1~A$) and compute voltage $V_{test}$.
• If the circuit does not contain any dependent sources, $R_{N}$ can be computed by deactivating all independent sources and looking at the resistance seen from the output terminals (in this case, $R_{N}$ can often be computed using series and parallel simplifications of resistors).
• If $I_N$ and $V_{Th}$ are known and $I_{N} \neq 0$, one can calculate the Norton resistance using $R_{N} = \frac{V_{Th}}{I_{N}}$. However, please note that if $I_{N}=0$, we cannot divide $\frac{V_{Th}}{I_{N}}$; this does not mean that $R_{N}$ is equal to $0$ (or $\infty$), but it only means we need to use other methods to compute $R_{N}$.

Table 5.2 shows possible ways to compute $R_{Th}$ and $I_{N}$. However, please note that these are not the only ways to compute the Norton components, and you can often come up with alternative ways.

¹⁾ You can use either the test voltage or test current method. They should both give you the same result.
²⁾ To deactivate the independent sources in the circuit, you need to replace all the independent voltage sources with short circuits (wires) and all the independent current sources with open circuits (remove them). Make sure you do not modify the dependent sources.

• Sometimes, the Norton voltage and Norton resistance can be computed simultaneously by performing successive source transformations and using series and parallel combinations of resistors, voltage sources and current sources.
• The Noton current of a circuit that does not contain any independent voltage and current sources is always 0.
• The Thévenin and Norton resistances, the Thévenin voltage and the Norton current satisfy the following relationships: $$\begin{equation}R_{Th}=R_N\end{equation}$$ $$\begin{equation}V_{Th}=I_N R_{Th}\end{equation}$$Because of these relationships, it is usually necessary to find only two quantities because the other two can be calculated afterwards.
• To deactivate a current source we remove it from the circuit (replace it with open-circuit or break); to deactivate a voltage source we replace it with a short-circuit (or wire).
• The above techniques can also be used to compute the Norton current and Norton impedance $Z_{N}$ in linear AC circuits.

Thévenin's theorem states that any two-terminal linear circuit containing only voltage sources, current sources and resistors (or impedances in the case of AC circuits) can be replaced by an equivalent combination of a voltage source $V_{Th}$ in series with a resistor $R_{Th}$ (or impedance $Z_{Th}$ in the case of AC circuits).

Thévenin's theorem and its dual, Norton's theorem, are widely used to simplify circuit analysis and study a circuit's initial-condition and steady-state response. Thévenin's theorem may in some cases be more convenient to use than Kirchhoff's circuit laws when analyzing linear circuits.

Calculating the Thévenin equivalent circuit, or simply the Thévenin equivalent of a circuit with two terminals means computing the Thévenin voltage and Thévenin resistance of the circuit.