ECE 211 · Lecture 4: Circuit Theorems & Equivalents

§1 · The theorem toolbox

Linearity underwrites everything

Six theorems make up the toolbox of this lecture, and all of them rest on one property of resistive circuits: linearity. The sixth entry builds directly on the Thevenin equivalent and is developed in the next lecture.

1 · Linearity 2 · Superposition 3 · Thevenin's theorem 4 · Norton's theorem 5 · Source transformation 6 · Maximum power transfer

A circuit is linear when its input-output relation satisfies two clauses:

Homogeneity (scaling). If the input is multiplied by $K$ , then the output is multiplied by $K$ .
Additivity. If the input is $V_1 + V_2$ , the output is $V_{o1} + V_{o2}$ , where $V_{o1}$ is caused by $V_1$ and $V_{o2}$ is caused by $V_2$ .

Both clauses are demonstrated below on the simplest possible circuit, a source driving one resistor. The sliders drive the build-time formulas, and the moving dots represent the resulting current.

Homogeneity: scaling the source by [K]{ln.ksrc} scales the current [I]{ln.ki} by the same K.

Additivity: with the input [V₁ + V₂]{ln.vs1 ln.vs2}, the current [I]{ln.ai} is the sum of the two partial currents.

K 1.5

V₁ (V) 4

V₂ (V) 6

I = K · 10 V5 Ω = 3.00 A (homogeneity)

I = 4 V5 Ω + 6 V5 Ω = 2.00 A (additivity)

Why resistive circuits qualify

Every governing equation of a resistive circuit is linear: Ohm's law is a proportionality, and KCL and KVL are sums. Power is the exception, because $P = v^2/R = i^2 R$ is quadratic; this exception returns as a warning in the next section.

§2 · Superposition

One source at a time

If a circuit has two or more independent sources, the output is the sum of the partial outputs caused by each source acting alone. The recipe from the board notes:

Turn OFF all but one independent source. A current source turned off becomes an open circuit; a voltage source turned off becomes a short circuit. Then solve the partial circuit.
Repeat for every independent source.
Sum the partial outputs: $V_o = V_{o1} + V_{o2} + V_{o3} + \cdots$

Turning a source off means setting its value to zero. A voltage source at 0 V holds no voltage across its terminals, which is a short; a current source at 0 A passes no current, which is an open.

Leave dependent sources in

Only independent sources are turned off. A dependent source is not an input; it is part of the circuit's internal behavior, and it must remain active in every partial circuit. The board notes carry this warning in red for good reason.

Power does not superpose

Superposition applies to voltages and currents, which are linear in the inputs. Power is quadratic, so the partial powers do not sum to the true power. Power must be computed from the total voltage and current, never from the partial circuits.

§3 · Worked example A: superposition

Find $V_o$ with superposition

The circuit below contains three independent sources and one dependent source, which is the worst case for the recipe: three partial circuits are required, and the dependent source rides along in every one of them.

The problem

Find $V_o$ , the voltage across the 8 Ω resistor, using superposition. Given: a 12 V source, a 4 V source, a 7 A source, and the dependent source $10I_b$ , controlled by the current $I_b$ in the 2 Ω resistor.

The plan: three partial circuits

Three independent sources produce three partial circuits, each retaining the dependent source. The partial outputs $V_{o1}$ , $V_{o2}$ , and $V_{o3}$ are then summed.

3 independent sources V_o = V_o1 + V_o2 + V_o3

Partial circuit 1: the 7 A source alone

Both voltage sources are replaced by shorts. KVL around the left loop then reads $2I_b = 0$ , so $I_b = 0$ , and the dependent source forces zero current: it behaves as an open branch. KCL at the output node therefore sends the entire 7 A upward through the 8 Ω resistor, which is $-7$ A in the direction of the $V_o$ polarity marks:

V_{o1} = \tm{sp.cs7}{(-7\ \text{A})} \cdot \tm{sp.r8}{(8\ \Omega)} = \boxed{-56\ \text{V}}

Partial circuit 2: the 12 V source alone

The 4 V source is shorted and the 7 A source is opened. KVL around the left loop gives $-12 + 2I_b = 0$ , so $I_b = 6$ A. The dependent source now pulls $10I_b = 60$ A out of the output node, and that current must arrive upward through the 8 Ω resistor:

V_{o2} = \tm{sp.dep}{(-10 I_b)} \cdot \tm{sp.r8}{(8\ \Omega)} = (-60)(8) = \boxed{-480\ \text{V}}

Partial circuit 3: the 4 V source alone

The 12 V source is shorted and the 7 A source is opened. KVL around the left loop gives $2I_b + 4 = 0$ , so $I_b = -2$ A; the sign simply means the actual current flows leftward. The same output relation then yields a positive partial output:

V_{o3} = \tm{sp.dep}{(-10 I_b)} \cdot \tm{sp.r8}{(8\ \Omega)} = (+20)(8) = \boxed{+160\ \text{V}}

Sum the partial outputs

V_o = V_{o1} + V_{o2} + V_{o3} = -56 - 480 + 160 \;\Longrightarrow\; \boxed{V_o = -376\ \text{V}}

The controlling current superposes as well: $I_b = 0 + 6 - 2 = 4$ A in the full circuit. Every linear quantity in the circuit is the sum of its partial values.

Check: a direct nodal solution

Both voltage sources are tied to the bottom rail, so two node voltages are known immediately: 12 V and 4 V. Ohm's law across the 2 Ω resistor gives $I_b = (12 - 4)/2 = 4$ A, in agreement with the superposed value. KCL at the output node (currents out) then reads

10 I_b + 7 + \frac{V_o}{8} = 0 \;\Longrightarrow\; V_o = -8\,(40 + 7) = -376\ \text{V}\ \checkmark

One equation confirms the three-circuit computation. Superposition is rarely the cheapest route; its value is conceptual, and it becomes indispensable when the sources differ in kind (for example, DC together with AC).

§4 · Thevenin and Norton

Two-terminal circuits collapse to two elements

Thevenin's theorem. Any linear two-terminal circuit can be replaced with a voltage source equal to the open-circuit voltage, in series with a resistor equal to the resistance looking in with all independent sources off.

Norton's theorem. Any linear two-terminal circuit can be replaced with a current source equal to the short-circuit current, in parallel with a resistor equal to the same looking-in resistance.

Thevenin: V_T in series with R_T. With the port open, no current flows, so V_oc = V_T.

Norton: I_N in parallel with R_N. With the port shorted, all of I_N takes the short, so I_sc = I_N.

The two equivalents describe the same circuit, so their parameters are locked together by $V_T = I_N R_T$ and $R_T = R_N$ .

One straight line describes the port

Everything the external world can learn about a linear two-terminal circuit is contained in the relation between its port voltage and its port current, and linearity makes that relation a straight line:

\tm{iv.ch}{i = \frac{V_T - v}{R_T}}

The three port experiments are three readings of this one line. Opening the port sets $i = 0$ and reads the intercept on the voltage axis, $V_{oc} = V_T$ ; shorting the port sets $v = 0$ and reads the intercept on the current axis, $I_{sc} = I_N$ ; turning the independent sources off collapses the line onto the origin and leaves only its slope, $-1/R_T$ . Two networks are equivalent precisely when they draw the same line, which is the reason the Thevenin and the Norton forms may be exchanged at will.

Connecting a resistor across the port adds a second line, $i = v/R_L$ , through the origin. The circuit settles at the one point that satisfies both, and as the load runs from a short to an open that point travels the whole characteristic from one intercept to the other. No load can move it off the line.

The port characteristic of a circuit with $V_T = 12$ V and $R_T = 4$ Ω, which the load cannot alter, crossed by the load line of the resistor connected across it.

R_L (Ω) 4

i = 12 V4 Ω + 4 Ω = 1.50 A v = i · 4 Ω = 6.00 V

port characteristicload line of the connected resistoroperating point

Which point on the line is the most profitable one to occupy is the question answered by the maximum power transfer theorem in Lecture 5.

Extracting the parameters

Three port experiments extract them:

With the terminals open, the port current is zero, so no voltage drops across $R_T$ and the port voltage equals the internal source: $V_{oc} = V_T$ . Measuring (or computing) the open-circuit voltage delivers the Thevenin source directly.

With the terminals shorted, the entire Norton current takes the short, so $I_{sc} = I_N$ ; in the Thevenin picture the same experiment gives $I_{sc} = V_T / R_T$ . The two experiments together therefore determine the resistance as well: $R_T = V_{oc} / I_{sc}$ .

With all independent sources in the circuit turned off (voltage sources shorted, current sources opened), the circuit is purely resistive, and the resistance looking in through the terminals is $R_{eq} = R_T = R_N$ . This is usually the fastest route to the resistance when no dependent sources are present.

V_T = V_oc I_N = I_sc R_T = R_N = R_eq V_T = I_N · R_T

§5 · Worked example B: both equivalents

Find the Norton and Thevenin equivalents

One circuit is reduced both ways. The open-circuit voltage is found by nodal analysis (Lecture 3 pays off immediately), the resistance is found by turning the sources off, and the short-circuit current is found from a second, nearly identical nodal solution. The two equivalents must agree, which provides the check.

The problem

Find the Thevenin equivalent of the circuit at the terminals a and b. Given: a 40 V source, a 5 A source, two 6 Ω resistors, and two 3 Ω resistors.

Count the unknowns

With the bottom rail as reference, the 40 V source pins its top node, and two node voltages remain: $V_x$ and $V_T$ . Because the port is open, no current leaves through terminal a, and the node voltage $V_T$ is the open-circuit voltage.

4 nodes − 1 reference − 1 voltage source = 2 unknowns: Vₓ, V_T

Nodal analysis delivers $V_{oc}$

\text{@}V_x:\quad \tm{tx.r6t}{\frac{V_x - 40}{6}} \tm{tx.cs5}{\;-\; 5} + \tm{tx.r6v}{\frac{V_x - 0}{6}} + \tm{tx.r3h}{\frac{V_x - V_T}{3}} = 0

\text{@}V_T:\quad \tm{tx.r3h}{\frac{V_T - V_x}{3}} + \tm{tx.r3v}{\frac{V_T - 0}{3}} \tm{tx.cs5}{\;+\; 5} = 0

Clearing denominators gives $4V_x - 2V_T = 70$ and $2V_T - V_x = -15$ . Substituting $V_x = 2V_T + 15$ into the first equation yields $6V_T = 10$ :

\boxed{V_T = \tfrac{5}{3}\ \text{V}} \qquad \left(V_x = \tfrac{55}{3} \approx 18.33\ \text{V}\right)

$R_T$ : turn off the independent sources

The 40 V source becomes a short and the 5 A source becomes an open. What remains is a pure resistor network seen from the terminals. The shorted source places the two 6 Ω resistors in parallel between $V_x$ and the bottom rail.

Combine the parallel pair

6 \parallel 6 = \tm{tx.r66}{3\ \Omega}

The path from terminal a now runs through 3 Ω in series with the new 3 Ω, all in parallel with the remaining 3 Ω.

Reduce to one resistance

R_T = 3 \parallel (3 + 3) = 3 \parallel 6 = \frac{3 \cdot 6}{3 + 6} = \boxed{2\ \Omega}

The Thevenin circuit

A $\tfrac{5}{3}$ V source in series with 2 Ω is indistinguishable from the original five-element circuit at the terminals: same open-circuit voltage, same resistance, same behavior under any load.

The Norton route: short the port

Short the terminals and count again

A short is placed across a-b, and the current $I_N$ through it is sought. The short merges the former $V_T$ node with the reference, so the node is relabeled O, at 0 V, and only one unknown remains:

3 nodes − 1 reference − 1 voltage source = 1 unknown: Vₓ

KCL at $V_x$

\tm{nx.r6t}{\frac{V_x - 40}{6}} + \tm{nx.r6v}{\frac{V_x - 0}{6}} + \tm{nx.r3h}{\frac{V_x - 0}{3}} \tm{nx.cs5}{\;-\; 5} = 0 \;\Longrightarrow\; 4V_x = 70 \;\Longrightarrow\; \boxed{V_x = \tfrac{35}{2}\ \text{V}}

KCL at O recovers $I_N$

\tm{nx.r3h}{\frac{0 - V_x}{3}} \tm{nx.cs5}{\;+\; 5} + \tm{nx.r3v}{\frac{0 - 0}{3}} + \tm{nx.in}{I_N} = 0 \;\Longrightarrow\; I_N = \frac{35}{6} - 5 \;\Longrightarrow\; \boxed{I_N = \tfrac{5}{6}\ \text{A}}

The 3 Ω resistor at the port carries nothing: both of its ends sit at 0 V, so the shorted branch steals its entire current.

The Norton circuit, and the check

A $\tfrac{5}{6}$ A source in parallel with 2 Ω (the same looking-in resistance as before). The two equivalents must satisfy $V_T = I_N R_N$ :

I_N R_N = \tfrac{5}{6} \cdot 2 = \tfrac{5}{3}\ \text{V} = V_T\ \checkmark

§6 · Source transformation

Trading one equivalent for the other

Because the Thevenin and Norton forms describe the same terminal behavior, a voltage source $V_s$ in series with $R$ may be exchanged for a current source $I_s$ in parallel with the same $R$ , in either direction, whenever

V_s = I_s R

The open-circuit voltages match ( $V = V_s$ on the left, $V = I_s R$ on the right), and so do the short-circuit currents. The sliders drive both circuits at once:

[V_s]{st.vs} in series with [R]{st.ra}

[I_s = V_s / R]{st.is} in parallel with [R]{st.rb}

V_s (V) 10

R (Ω) 4

I_s = V_sR = 10 V4 Ω = 2.50 A

Polarity convention

The current-source arrow points toward the terminal that carries the voltage source's + mark. Reversing either one flips the sign of the equivalent, which is the most common source-transformation error.

§7 · Worked example C: a transformation chain

Find $V_x$ by repeated transformation

Nothing in this circuit is in series or in parallel from the viewpoint of the port, yet no nodal system is needed: a chain of source transformations collapses the circuit around the terminals of interest. The port carrying $V_x$ is preserved at every step, which is the one rule that must never be broken.

The problem

Find $V_x$ , the voltage across the 5 A source. Given: an 8 Ω resistor in parallel with the source, a 30 Ω outer branch, a 40 V source in series with 20 Ω, and a 12 Ω branch on the right.

Transform at both ends

The 40 V + 20 Ω series branch becomes a current source in parallel with 20 Ω, and the 5 A ∥ 8 Ω pair becomes a voltage source in series with 8 Ω. The port terminals are untouched.

I = \frac{40}{20} = 2\ \text{A (arrow toward the + side)} \qquad\quad V = 5 \cdot 8 = 40\ \text{V}

Combine the parallel resistors

The 20 Ω now shares both nodes with the 30 Ω outer branch:

20 \parallel 30 = \frac{20 \cdot 30}{20 + 30} = \tm{sx.r1230}{12\ \Omega}

Transform back to a voltage source

\tm{sx.cs2}{2\ \text{A}} \parallel 12\ \Omega \;\longrightarrow\; \tm{sx.vs24}{V = 2 \cdot 12 = 24\ \text{V}} \ \text{in series with}\ 12\ \Omega

Combine the series resistors

The two 12 Ω resistors now carry the same current:

12 + 12 = \tm{sx.r24}{24\ \Omega}

Transform both series branches

Each voltage-plus-resistor branch hanging on the port becomes a current source in parallel with its resistor, and both arrows point toward the top rail:

\frac{40}{8} = \tm{sx.cs5b}{5\ \text{A}} \qquad\quad \frac{24}{24} = \tm{sx.cs1}{1\ \text{A}}

Combine and read off the answer

Parallel current sources add, and the parallel resistors combine:

5 + 1 = \tm{sx.cs6}{6\ \text{A}} \qquad 8 \parallel 24 = \tm{sx.r6}{6\ \Omega} \qquad V_x = 6 \cdot 6 \;\Longrightarrow\; \boxed{V_x = 36\ \text{V}}

With the port open, the entire 6 A flows through the 6 Ω resistor, and the port voltage follows from Ohm's law.

Check: nodal analysis of the original circuit

With the bottom rail as reference, $V_x$ names the top-left node and $w$ the node joining the 30 Ω, 20 Ω, and 12 Ω branches. KCL at the two nodes:

\text{@}V_x:\quad -5 + \frac{V_x}{8} + \frac{V_x - 40 - w}{20} + \frac{V_x - w}{30} = 0

\text{@}w:\quad \frac{w - V_x}{30} + \frac{w - (V_x - 40)}{20} + \frac{w}{12} = 0

Clearing denominators gives $5V_x - 2w = 168$ and $10w = 5V_x - 120$ ; solving the pair yields $w = 6$ V and $V_x = 36$ V. Substituting back: $-5 + 4.5 - 0.5 + 1 = 0$ ✓ and $-1 + 0.5 + 0.5 = 0$ ✓. The transformation chain and the nodal solution agree.

§8 · Your turn

Practice: one circuit, three routes

The circuit below is small, but it exercises the entire lecture: the Thevenin and Norton equivalents at the terminals a-b can be reached by nodal analysis, by superposition, or by a transformation chain, and all three routes must agree. The problem should be attempted on paper before the hints are opened, in order.

Find the Thevenin and Norton equivalents at the terminals a and b. Given: a 36 V source in series with 12 Ω, a 3 A source, and a 6 Ω resistor.

Hint 1: The resistance costs nothing

With the 36 V source shorted and the 3 A source opened, the 12 Ω and 6 Ω resistors are in parallel between the terminals: $R_T = 12 \parallel 6 = 4\ \Omega$ .

Hint 2: Three routes to V_oc

Route 1 (nodal): one KCL equation at the top node delivers $V_{oc}$ directly. Route 2 (superposition): the 36 V source alone drives a voltage divider, and the 3 A source alone drives the parallel resistance. Route 3 (transformation): converting the 36 V + 12 Ω branch into a current source turns the whole circuit into parallel elements.

Hint 3: The equations

Nodal: $\tm{pr.r12}{\frac{V - 36}{12}} + \tm{pr.r6}{\frac{V - 0}{6}} = \tm{pr.cs3}{3}$ . Superposition: $36 \cdot \frac{6}{12 + 6} + 3 \cdot (12 \parallel 6)$ . Transformation: $\frac{36}{12} = 3$ A, then $(3 + 3)\ \text{A}$ into $12 \parallel 6$ .

Solution

All three routes give the same open-circuit voltage. Nodal: multiplying by 12 gives $V - 36 + 2V = 36$ , so $3V = 72$ and $V_{oc} = 24$ V. Superposition: $36 \cdot \tfrac{1}{3} + 3 \cdot 4 = 12 + 12 = 24$ V. Transformation: $6\ \text{A} \cdot 4\ \Omega = 24$ V.

\boxed{V_T = 24\ \text{V}, \quad R_T = 4\ \Omega} \qquad\quad \boxed{I_N = \tfrac{V_T}{R_T} = 6\ \text{A}, \quad R_N = 4\ \Omega}

The short-circuit experiment confirms $I_N$ independently: with the port shorted, the top node is pinned at 0 V, so the 36 V source pushes $36/12 = 3$ A through the 12 Ω resistor, the 6 Ω resistor carries nothing, and the 3 A source adds directly: $I_{sc} = 3 + 3 = 6$ A. ✓

Linearity underwrites everything

One source at a time

Find VoV_oVo​ with superposition

The problem

The plan: three partial circuits

Partial circuit 1: the 7 A source alone

Partial circuit 2: the 12 V source alone

Partial circuit 3: the 4 V source alone

Sum the partial outputs

Two-terminal circuits collapse to two elements

One straight line describes the port

Extracting the parameters

Find the Norton and Thevenin equivalents

The problem

Count the unknowns

Nodal analysis delivers VocV_{oc}Voc​

RTR_TRT​: turn off the independent sources

Combine the parallel pair

Reduce to one resistance

The Thevenin circuit

The Norton route: short the port

Short the terminals and count again

KCL at VxV_xVx​

KCL at O recovers INI_NIN​

The Norton circuit, and the check

Trading one equivalent for the other

Find VxV_xVx​ by repeated transformation

The problem

Transform at both ends

Combine the parallel resistors

Transform back to a voltage source

Combine the series resistors

Transform both series branches

Combine and read off the answer

Practice: one circuit, three routes

Find $V_o$ with superposition

Nodal analysis delivers $V_{oc}$

$R_T$ : turn off the independent sources

KCL at $V_x$

KCL at O recovers $I_N$

Find $V_x$ by repeated transformation