Use Thevenin equivalent circuits: Exercises — Hardware deep dive

Throughout, we use these symbols, each defined before use:

== $V_{t h}$ == — the open-circuit voltage: the voltage you would read with a perfect voltmeter across the two terminals when nothing is connected to them.
== $R_{t h}$ == — the Thevenin resistance: the resistance you would measure looking into the two terminals after making every independent source dead (voltage source → replaced by a wire, current source → replaced by a gap).
== $I_{sc}$ == — the short-circuit current: the current that flows if you join the two terminals with a plain wire.
The terminal law every Thevenin box obeys: $V = V_{t h} - I R_{t h}$ , a straight line whose intercept is $V_{t h}$ and whose slope is $- R_{t h}$ .

The parallel-resistance shorthand $R_{1} ∥ R_{2}$ means $\frac{R _{1} R _{2}}{R _{1} + R _{2}}$ — the resistance of two resistors sharing the same two end-nodes.

Figure — Use Thevenin equivalent circuits

Level 1 — Recognition

L1.1

A two-terminal linear box is measured. Open-circuit voltage is $9 V$ ; when the terminals are shorted, $3 A$ flows. What are $V_{t h}$ and $R_{t h}$ ?

Recall Solution

WHAT we do: read the two numbers straight off the definitions. $V_{t h}$ is the open-circuit voltage, so $V_{t h} = 9 V$ . For $R_{t h}$ , use the short-circuit relation $R_{t h} = V_{t h} / I_{sc}$ (from the terminal law with $V = 0$ ): $R_{t h} = \frac{9}{3} = 3 Ω.$ Answer: $V_{t h} = 9 V$ , $R_{t h} = 3 Ω$ .

L1.2

A Thevenin box has $V_{t h} = 10 V$ , $R_{t h} = 5 Ω$ . What current flows if a $5 Ω$ load is connected? What is the terminal voltage then?

Recall Solution

WHY the terminal law: the load and the box share one loop, so the same current $I$ runs through both. Apply $V = V_{t h} - I R_{t h}$ together with Ohm's law on the load, $V = I R_{L}$ . $I R_{L} = V_{t h} - I R_{t h} \Rightarrow I = \frac{V _{t h}}{R _{t h} + R _{L}} = \frac{10}{5 + 5} = 1 A .$ Terminal voltage: $V = I R_{L} = 1 \times 5 = 5 V$ . Answer: $I = 1 A$ , $V = 5 V$ .

Level 2 — Application

L2.1

A source $V_{S} = 24 V$ feeds $R_{1} = 6 Ω$ (series) then $R_{2} = 3 Ω$ (to ground). The output terminals are across $R_{2}$ . Find $V_{t h}$ and $R_{t h}$ .

Recall Solution

Step 1 — $V_{t h}$ (OPEN for voltage). Load removed, so $R_{1}$ and $R_{2}$ carry the same current and form a plain voltage divider: $V_{t h} = V_{S} \frac{R _{2}}{R _{1} + R _{2}} = 24 \cdot \frac{3}{6 + 3} = 24 \cdot \frac{1}{3} = 8 V .$ Step 2 — $R_{t h}$ (DEAD for resistance). Short $V_{S}$ . Its top node becomes ground, so $R_{1}$ now runs from the terminal node to ground alongside $R_{2}$ — they are in parallel: $R_{t h} = R_{1} ∥ R_{2} = \frac{6 \cdot 3}{6 + 3} = \frac{18}{9} = 2 Ω.$ Answer: $V_{t h} = 8 V$ , $R_{t h} = 2 Ω$ .

L2.2

For the box in L2.1, connect a load $R_{L} = 4 Ω$ . Find the load current and the power in the load.

Recall Solution

$I_{L} = \frac{V _{t h}}{R _{t h} + R _{L}} = \frac{8}{2 + 4} = \frac{8}{6} = 1.333 A .$ Power: $P_{L} = I_{L}^{2} R_{L} = (1.333)^{2} \times 4 = 7.111 W$ . Answer: $I_{L} \approx 1.333 A$ , $P_{L} \approx 7.11 W$ .

L2.3

Confirm $R_{t h}$ of L2.1 a second way, by short-circuit current. Short the output terminals of the original circuit and find $I_{sc}$ , then $R_{t h} = V_{t h} / I_{sc}$ .

Recall Solution

WHAT the short does: joining the terminals ties the $R_{2}$ node to ground, so $R_{2}$ has $0 V$ across it and carries no current — all the source current goes through $R_{1}$ into the short. $I_{sc} = \frac{V _{S}}{R _{1}} = \frac{24}{6} = 4 A, R_{t h} = \frac{V _{t h}}{I _{sc}} = \frac{8}{4} = 2 Ω. ✅$ Matches L2.1.

Level 3 — Analysis

L3.1

Two independent sources: a $12 V$ source through $R_{1} = 4 Ω$ into node A, and a $6 V$ source through $R_{2} = 2 Ω$ into the same node A. From A a resistor $R_{3} = 6 Ω$ goes to ground. Terminals are across $R_{3}$ . Find $V_{t h}$ and $R_{t h}$ .

Recall Solution

Step 1 — $V_{t h}$ by node analysis at A (open circuit). With the load off, $R_{3}$ is still in the circuit as the path to ground. Let $V_{A}$ be the node voltage. Sum currents leaving A: $\frac{V _{A} - 12}{4} + \frac{V _{A} - 6}{2} + \frac{V _{A} - 0}{6} = 0.$ Multiply by 12: $3 (V_{A} - 12) + 6 (V_{A} - 6) + 2 V_{A} = 0 \Rightarrow 11 V_{A} - 72 = 0 \Rightarrow V_{A} = \frac{72}{11} = 6.545 V .$ The open-circuit terminal voltage is $V_{A}$ , so $V_{t h} = 6.545 V$ . Step 2 — $R_{t h}$ (both sources dead). Short both voltage sources. Now $R_{1}$ , $R_{2}$ , $R_{3}$ all run from node A to ground — three resistors in parallel: $R_{t h} = (\frac{1}{4} + \frac{1}{2} + \frac{1}{6})^{- 1} = (\frac{3 + 6 + 2}{12})^{- 1} = \frac{12}{11} = 1.091 Ω.$ Answer: $V_{t h} = 6.545 V$ , $R_{t h} = 1.091 Ω$ .

L3.2

For the box in L3.1, verify $V_{t h}$ using superposition (each source acting alone).

Recall Solution

Idea: with the load off, the terminal node A sees a divider each time. 12 V alone (short the 6 V source): $R_{2} ∥ R_{3} = 2∥6 = 1.5 Ω$ hangs from A to ground, fed through $R_{1} = 4$ : $V_{A}^{(12)} = 12 \cdot \frac{1.5}{4 + 1.5} = 12 \cdot \frac{1.5}{5.5} = 3.273 V .$ 6 V alone (short the 12 V source): $R_{1} ∥ R_{3} = 4∥6 = 2.4 Ω$ from A to ground, fed through $R_{2} = 2$ : $V_{A}^{(6)} = 6 \cdot \frac{2.4}{2 + 2.4} = 6 \cdot \frac{2.4}{4.4} = 3.273 V .$ Add: $V_{t h} = 3.273 + 3.273 = 6.545 V . ✅$ Matches L3.1.

Level 4 — Synthesis

L4.1

Convert the Thevenin box from L2.1 ( $V_{t h} = 8 V$ , $R_{t h} = 2 Ω$ ) into its Norton equivalent via source transformation. State $I_{N}$ and $R_{N}$ .

Recall Solution

Rule: a Thevenin source ( $V_{t h}$ in series with $R_{t h}$ ) equals a Norton source ( $I_{N}$ in parallel with $R_{N}$ ) when $R_{N} = R_{t h}$ and $I_{N} = V_{t h} / R_{t h}$ (this $I_{N}$ is exactly the short-circuit current). $R_{N} = R_{t h} = 2 Ω, I_{N} = \frac{V _{t h}}{R _{t h}} = \frac{8}{2} = 4 A .$ Answer: $I_{N} = 4 A$ in parallel with $R_{N} = 2 Ω$ . (Note $I_{N} = I_{sc} = 4 A$ from L2.3 — consistent.)

L4.2

Using L2.1's box, apply the Maximum Power Transfer Theorem: what load $R_{L}$ draws maximum power, and what is that power?

Recall Solution

WHY $R_{L} = R_{t h}$ : power in the load is $P_{L} = (\frac{V _{t h}}{R _{t h} + R _{L}})^{2} R_{L}$ . Setting $d P_{L} / d R_{L} = 0$ gives $R_{L} = R_{t h}$ — the load matched to the source resistance. Here $R_{t h} = 2 Ω$ , so: $R_{L} = 2 Ω, P_{ma x} = \frac{V _{t h}^{2}}{4 R _{t h}} = \frac{8 ^{2}}{4 \cdot 2} = \frac{64}{8} = 8 W .$ Answer: $R_{L} = 2 Ω$ , $P_{ma x} = 8 W$ .

Level 5 — Mastery

L5.1 (dependent source)

A network has a $2 Ω$ resistor from the terminal node to ground, in parallel with a dependent current source that pushes $2 i_{x}$ (amps) from ground up to the terminal node, where $i_{x}$ is the current flowing down through that same $2 Ω$ resistor. There are no independent sources. Find $R_{t h}$ using the test-source method.

Recall Solution

WHY a test source: with dependent sources present you cannot deactivate them (they track a circuit variable). Instead drive the terminals with a known test voltage $V_{t}$ , find the resulting current $I_{t}$ into the terminal, and take $R_{t h} = V_{t} / I_{t}$ .

Let the terminal node sit at $V_{t}$ above ground. The $2 Ω$ resistor carries $i_{x} = V_{t} /2$ downward. The dependent source injects $2 i_{x}$ into the node.

KCL at the node (currents leaving = current we inject $I_{t}$ ): $I_{t} + 2 i_{x} = i_{x} \Rightarrow I_{t} = i_{x} - 2 i_{x} = - i_{x} = - \frac{V _{t}}{2} .$ Wait — carefully: current from the test source in plus the dependent source in must equal current out through the resistor: $I_{t} + 2 i_{x} = i_{x} \Rightarrow I_{t} = - i_{x} = - \frac{V _{t}}{2} .$ Then $R_{t h} = \frac{V _{t}}{I _{t}} = \frac{V _{t}}{- V _{t} /2} = - 2 Ω.$ Answer: $R_{t h} = - 2 Ω$ . A negative Thevenin resistance is legal for dependent-source networks — it means the box supplies energy rather than dissipating it as current is pushed in. This is exactly why you may never "deactivate" a dependent source: doing so would have given the wrong $+ 2 Ω$ .

L5.2 (full chain)

Back to L3.1's box ( $V_{t h} = 6.545 V$ , $R_{t h} = 1.091 Ω$ ). (a) What load gives maximum power and how much? (b) If instead $R_{L} = 3 Ω$ , find the load current and terminal voltage.

Recall Solution

(a) Matched load $R_{L} = R_{t h} = 1.091 Ω$ . $P_{ma x} = \frac{V _{t h}^{2}}{4 R _{t h}} = \frac{( 6.545 ) ^{2}}{4 \cdot 1.091} = \frac{42.84}{4.364} = 9.818 W .$ (b) With $R_{L} = 3 Ω$ : $I_{L} = \frac{V _{t h}}{R _{t h} + R _{L}} = \frac{6.545}{1.091 + 3} = \frac{6.545}{4.091} = 1.600 A .$ Terminal voltage: $V = I_{L} R_{L} = 1.600 \times 3 = 4.800 V$ (check against terminal law: $6.545 - 1.600 \cdot 1.091 = 4.800 V$ ✅). Answer: (a) $R_{L} = 1.091 Ω$ , $P_{ma x} = 9.82 W$ ; (b) $I_{L} = 1.600 A$ , $V = 4.80 V$ .

Connections

Voltage and Current Dividers — the workhorse for every $V_{t h}$ divider here.
Nodal and Mesh Analysis — used in L3 for multi-source $V_{t h}$ .
Superposition Theorem — the L3.2 cross-check.
Source Transformation & Norton Equivalent Circuits — the L4.1 conversion.
Maximum Power Transfer Theorem — L4.2 and L5.2.
Back to parent: Use Thevenin equivalent circuits.

Exercises — Use Thevenin equivalent circuits

Level 1 — Recognition

L1.1

L1.2

Level 2 — Application

L2.1

L2.2

L2.3

Level 3 — Analysis

L3.1

L3.2

Level 4 — Synthesis

L4.1

L4.2

Level 5 — Mastery

L5.1 (dependent source)

L5.2 (full chain)

Connections

Difficulty Map