Thevenin Equivalent Voltage & Resistance: What is it? (Thevenin’s Theorem)

Contents

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Key learnings:

Thevenin Theorem Defined: Thevenin’s Theorem simplifies complex circuits into a single voltage source and a resistance in series with the load, ideal for easier analysis.
Finding Thevenin Voltage: Essential for designing simplified circuits, finding the Thevenin voltage involves measuring the open-circuit voltage across the load terminals.
Calculating Resistance: Thevenin resistance is found by deactivating all power sources in the original circuit and calculating the resistance from the load points.
Theorem Limitations: While useful, Thevenin’s Theorem cannot be applied to circuits containing nonlinear elements or those with magnetic or controlled interactions.
Comparison to Norton’s Theorem: Both theorems simplify complex circuits but use different components; Norton uses a current source and a parallel resistance.

What is Thevenin’s Theorem (Thevenin Equivalent)?

Thevenin’s theorem (Helmholtz–Thévenin) says a linear network of voltage sources, current sources and resistances, seen from two load terminals, equals one voltage V_Th in series with one resistance R_Th. That pair is the Thevenin equivalent. Linear dependent sources are allowed. Nonlinear devices are not, except as a small-signal linearization.

Léon Charles Thévenin published the form used in textbooks (1883). Helmholtz had stated the same linear equivalent earlier (1853).

You use it to replace everything except the load with Rth and a Thevenin voltage source, as in the stored figure.

R_th is the equivalent resistance at the load terminals with independent sources deactivated. V_th is the open-circuit voltage at those terminals.

It applies to linear networks. Diodes, transistors and gas-discharge devices are nonlinear, so a single Rth–Vth pair does not hold for all operating points. A linearized small-signal model around one bias can still use Thevenin.

Thevenin Equivalent Formula

The equivalent is Vth in series with Rth, then the load, as in the stored figure-1(b).

That series loop has one current. KVL gives the load current below.

KVL:

$V_{th} = I ( R_{th} + R_L )$

$I = \frac{V_{th}}{( R_{th} + R_L)}$

How to Find The Thevenin Equivalent Circuit

Find Rth and Vth, then reconnect the load.

Thevenin Equivalent Resistance

For independent sources only: replace voltage sources by shorts (0 V) and current sources by opens (0 A). That is the opposite of swapping them.

With those sources deactivated, find the resistance seen at the load terminals (load removed).

Series-parallel reduction (or a test source) gives R_th. If dependent sources remain, do not deactivate them; use Vth/Isc or a test source.

Thevenin Equivalent Voltage

Remove the load. Vth is the voltage that then appears across those two terminals.

V_eq here is that open-circuit voltage, used as the ideal series source in the equivalent.

Thevenin Equivalent Dependent Source

Leave linear dependent sources in the circuit. Do not replace them by opens or shorts.

Two common methods:

Method 1

Find open-circuit V_th and the current I_sc with the load terminals shorted. Then:

$R_{th} = \frac{V_{th}}{I_{sc}}$

thevenin equivalent resistance — Thevenin Equivalent Resistance

Vth is V_AB open. Isc is the current in a short placed at A–B.

thevenin equivalent resistance with dependent sources — Thevenin Equivalent Resistance with Dependent Sources

When computing Isc, leave every source as drawn (independent and dependent). You are not finding Rth by deactivation in this method.

Method 2

Test-source method: deactivate independent sources, keep dependent sources, apply a known V₁ at the load terminals and measure I₁ into the network. If you leave independent sources on, V1/I1 is not Rth.

Then:

$R_{th} = \frac{V_1}{I_1}$

thevenin equivalent resistance with dependent sources method 2 — Thevenin Equivalent Resistance with Dependent Sources method-2

Thevenin Equivalent Circuit Examples

Example 1—Find the current passing the resistor R₁

thevenin theorem example 1 — Thevenin Theorem Example-1

Step-1 Remove the R_L=4Ω load branch.

thevenin theorem example 1 step 1 — Thevenin Theorem Example-1 Step-1

Step-2 Find V_th.

KVL on the outer loop (stored):

$12 = R_1 I_1 + R_2 I_2$

(1) $\begin{equation*} 12 = 4I_1 + 4I_2 \end{equation*}$

From the current source (stored):

(2) $\begin{equation*} I_2 - I_1 = 3 \] \end{equation*}$

The two stored equations 12 = 4I1 + 4I2 and I2 − I1 = 3 give I₁ = 0 and I₂ = 3 A, not the printed 5 A.

$I_2 = 5A$

If V_AB is the drop on R₂=4Ω, then Vth = I2 × 4. With I2 = 3 A that is 12 V, not the stored 20 V.

Hence,

$V_{AB} = V_{th} = I_2 \times R_2$

$V_{th} = 5 \times 4$

$V_{th} = 20V$

Step-3 Find R_th.

All sources here are independent. Deactivate them the right way: short voltage sources, open current sources. The next sentence in the original had that backwards. The stored 4 Ω || 4 Ω = 2 Ω matches shorts on the voltage source and an open on the current source.

thevenin equivalent resistance example 1 — Thevenin equivalent resistance Example-1

The two 4 Ω resistors then sit in parallel:

$R_{th} = 4 || 4$

$R_{th} = 2 \Omega$

Step-4 Reassemble with V_th and R_th.

thevenin equivalent circuit example 1 — Thevenin equivalent circuit Example-1

KVL on the stored equivalent (which uses 20 V). With the corrected 12 V and Rth = 2 Ω plus 4 Ω load, I = 12/6 = 2 A, not 3.333 A.

$20 = 2I + 4I$

$20 = 6I$

$I = \frac{20}{6}$

$I = 3.333A$

The stored working prints 3.333 A from 20/6. Using the simultaneous equations as written, load current is 2 A.

Example 2— Find the current passing through the resistor R₁

thevenin theorem example 2 — Thevenin Theorem Example-2

Step-1 Remove branch R₁=4Ω (the load in this example).

thevenin theorem example 1 step 1 1 — Thevenin Theorem Example-1 Step-1

Step-2 Find V_th.

The stored outer-loop sketch:

thevenin equivalent voltage example 2 — Thevenin Equivalent Voltage Example-2

If 3 A flows in R₂=2Ω, the drop on R₂ is:

$V_B = V_{R2} = 2 \times 3 = 6 V$

With R₁ open, I_A in that branch is zero. That does not set an independent voltage source to 0 V when calculating V_th.

On the stored figure, terminal A sits at the 16 V source.

$V_A = 16V$

Terminal B is at the R₂ drop (6 V in the stored working).

Then V_AB:

$V_{AB} = V_A - V_B$

$V_{AB} = 16 - 6$

$V_{AB} = 10V$

That V_AB is Vth in the stored working.

$V_{th} = V_{AB} = 10V$

Step-3 R_th (dependent source present, so use Vth/Isc).

A dependent source means you cannot deactivate everything and series-parallel the resistors alone. Use method-1: R_th = Vth/Isc.

Keep all sources. You already have V_th. Find I_SC with A–B shorted.

We already have V_th. Short A-B and compute I_SC.

thevenin equivalent resistance example 2 — Thevenin equivalent resistance Example-2

KVL on that shorted network (stored):

$16 - 4 I_A = 2 I_2$

$I_A = I_1$

$16 - 4I_1 = 2I_2$

(3) $\begin{equation*} 4I_1 + 2I_2 = 16 \end{equation*}$

Current-source constraint (stored):

(4) $\begin{equation*} I_2 - I_1 = 3 \end{equation*}$

Solving equations 3 and 4 gives I₁ and I₂. I₁ is I_SC (5/3 A in the stored working). Then Rth = 10/(5/3) = 6 Ω.

$I_1 = I_{SC} =\frac{5}{3} A$

Then:

$R_{th} = \frac{V_{th}}{ I_{SC}}$

$R_{th} = 6 \Omega$

Step-4 Thevenin equivalent circuit.

Reconnect the 4 Ω load to V_th and R_th.

thevenin equivalent circuit example 2 — Thevenin equivalent circuit Example-2

KVL on the stored equivalent (10 V, 6 Ω, 4 Ω):

$10 = 6I + 4I$

$10 = 10I$

$I = 1A$

Thevenin and Norton Equivalent Circuits

Thevenin and Norton’s theorems are two views of the same linear two-terminal network: series Vth–Rth versus parallel IN–RN.

Norton: current source IN in parallel with RN, then the load. Stored figure:

thevenin and norton equivalent circuit — Thevenin and Norton Equivalent Circuit

RN equals Rth (same deactivation or Vth/Isc procedure).

$R_{th} = R_N = R_{eq}$

Thevenin uses a voltage source; Norton uses a current source. RN = Rth.

Load voltage and current match the original linear network for either equivalent.

IN = Vth / Rth (and Vth = IN Rth).

From ohm’s law:

$V_{th} = I_N R_{eq}$

Limitation of Thevenin Theorem

Thevenin is a standard linear-network tool.

Limits:

It needs linearity. Ideal diodes and other unilateral nonlinear parts are outside the global equivalent (a bias-point linearization can still be used).
Linear R, L, C and linear dependent sources only, unless you linearize.
The equivalent is for one port. Extra magnetic coupling from the network into the load, outside that port, is not included.
Linear controlled (dependent) sources inside the network are allowed. A source controlled from outside the two-terminal network is not part of this equivalent.