1. Network Theory Basics and Overview of Theorems
Network theory (circuit analysis) studies how voltages and currents behave in a network of interconnected elements such as resistors, capacitors, inductors and sources. The basic tools are Ohm's law and Kirchhoff's laws (KCL and KVL). Solving large circuits with them alone means many simultaneous equations, so engineers use electrical network theorems as shortcuts.
Key Terms You Must Know
| Term | Meaning |
|---|---|
| Node / Branch / Loop / Mesh | Node: junction of two or more elements. Branch: one element between two nodes. Loop: any closed path. Mesh: a loop with no other loop inside it. |
| Active / Passive element | Active elements supply energy (sources). Passive elements (R, L, C) absorb or store it. |
| Linear network | Output is proportional to input; obeys superposition and homogeneity. Resistor values do not depend on voltage or current. |
| Bilateral element | Behaves the same for current in either direction (resistor, inductor, capacitor). Diodes and transistors are unilateral. |
| Independent source | Value fixed by itself. Deactivated by: voltage source → short circuit, current source → open circuit. |
| Dependent source | Value controlled by a voltage or current elsewhere in the circuit. It is never deactivated. |
| Lumped network | Elements are treated as points; no wave-propagation delay. KCL and KVL apply. |
Figure 1: The ten network theorems and the one-line idea behind each. Sections 2 to 10 explain each in detail.
Which Theorem Do I Use? (Quick Guide)
Several sources, want one response → superposition. Want current or voltage in one changing load → Thevenin or Norton. Want the best load for power → maximum power transfer. Parallel voltage-source branches → Millman. Check a solution → Tellegen (power balance). Effect of changing one resistor → compensation.
2. Superposition Theorem
Statement
In a linear network containing more than one independent source, the current through (or voltage across) any element equals the algebraic sum of the currents (or voltages) produced by each independent source acting alone, with all other independent sources deactivated.
Steps
- Keep one independent source active. Replace every other independent voltage source by a short circuit and every independent current source by an open circuit.
- Find the required current or voltage due to that source alone.
- Repeat for each independent source in turn.
- Add the results algebraically, paying attention to direction and polarity.
Limits of Superposition
- Applies only to linear circuits.
- Not applicable to power: power depends on the square of current or voltage, so P ≠ P₁ + P₂. Find total current or voltage first, then compute power.
- Dependent sources stay active in every step; only independent sources are deactivated.
Figure 2: Superposition. In (b) the current source is opened; in (c) the voltage source is replaced by a short circuit. The two partial answers add up to the real voltage at node A.
Check by Nodal Analysis
KCL at A: (VA − 12)/2 + VA/4 = 6 → 0.75 VA = 12 → VA = 16 V. It matches the superposition result, as it must.
3. Thevenin's Theorem
Statement
Any linear two-terminal network containing sources and resistances can be replaced by an equivalent circuit made of a single voltage source Vth in series with a single resistance Rth. Vth is the open-circuit voltage across the terminals; Rth is the resistance seen from the terminals with all independent sources deactivated.
The theorem is named after the French engineer Léon Charles Thévenin, who published it in 1883. It is mainly used to find the current in, or voltage across, one element (the load) that can change, without re-solving the whole network each time.
Steps to Apply Thevenin's Theorem
- Remove the load and mark the two terminals a and b.
- Find Vth = the open-circuit voltage between a and b.
- Find Rth: deactivate all independent sources (short voltage sources, open current sources) and calculate the resistance looking into a–b. If the network has dependent sources, connect a test source and use Rth = Vtest/Itest, or use Rth = Vth/Isc.
- Draw the Thevenin equivalent and reconnect the load.
- Load current: IL = Vth / (Rth + RL).
Thevenin Formulas
Vth = open-circuit voltage | Isc = short-circuit current at the same terminals
Figure 3: Thevenin's theorem. Looking into terminals a–b, the whole source-and-two-resistor network behaves like a 16 V source with 4 Ω in series.
Worked Result for a 4 Ω Load
Connect RL = 4 Ω across a–b: IL = 16 / (4 + 4) = 2 A, VL = 8 V. Direct check: 12 ∥ 4 = 3 Ω; total = 6 + 3 = 9 Ω; source current = 24/9 = 2.667 A; voltage across the parallel part = 2.667 × 3 = 8 V; load current = 8/4 = 2 A. ✓
4. Norton's Theorem and Source Transformation
Statement
Any linear two-terminal network can be replaced by an equivalent circuit made of a single current source IN in parallel with a single resistance RN. IN is the short-circuit current between the terminals, and RN equals the Thevenin resistance (RN = Rth).
Edward Lawry Norton, an engineer at Bell Telephone Laboratories, described this dual form in 1926. Norton's and Thevenin's equivalents are interchangeable.
Norton Formulas and Link with Thevenin
Load current from the Norton form (current divider): IL = IN · RN / (RN + RL)
Steps to Apply Norton's Theorem
- Remove the load and short the terminals a–b.
- Calculate the short-circuit current IN through the short.
- Find RN exactly as Rth: deactivate independent sources and look into a–b.
- Draw IN in parallel with RN, reconnect the load and apply the current-divider rule.
Norton Equivalent of the Same Circuit
With terminals a–b shorted in Figure 3, the 12 Ω resistor is bypassed, so IN = 24 / 6 = 4 A. RN = 4 Ω. Check: Vth = IN · RN = 4 × 4 = 16 V ✓. For RL = 4 Ω: IL = 4 × 4/(4 + 4) = 2 A ✓.
Source Transformation
Source transformation is the quick tool behind the Thevenin–Norton link: a voltage source Vs in series with Rs is equivalent, as seen from the outside terminals, to a current source Is = Vs/Rs in parallel with the same Rs.
Figure 4: Source transformation converts between the Thevenin form (voltage source plus series resistor) and the Norton form (current source plus parallel resistor).
Source Transformation Cautions
- Keep the polarity: the arrow of the current source points toward the + terminal of the original voltage source.
- An ideal voltage source with no series resistance (or an ideal current source with no parallel resistance) cannot be transformed.
- A resistor in parallel with a voltage source, or in series with a current source, does not affect the terminal behaviour and can be ignored when finding terminal quantities.
5. Maximum Power Transfer Theorem
Statement
A source delivers maximum power to a load when the load resistance equals the Thevenin resistance of the network feeding it: RL = Rth. (In AC circuits, the load impedance must equal the complex conjugate of the source impedance: ZL = Zth*.)
Maximum Power Transfer Formulas
Equivalent Norton form: Pmax = IN² RN / 4 | Efficiency at maximum power transfer: η = 50 %
Why RL = Rth?
Power in the load is P = I²RL with I = Vth/(Rth + RL). If RL is very small, the current is large but the voltage across the load is tiny. If RL is very large, the voltage is large but the current is tiny. Setting dP/dRL = 0 gives the best compromise at RL = Rth.
Figure 5: Load power rises to a single peak at RL = Rth and then falls, so there is exactly one best load resistance.
| RL (Ω) | IL = 16/(4+RL) (A) | PL = IL² RL (W) | Efficiency RL/(Rth+RL) |
|---|---|---|---|
| 1 | 3.20 | 10.24 | 20 % |
| 2 | 2.67 | 14.22 | 33 % |
| 4 | 2.00 | 16.00 (max) | 50 % |
| 8 | 1.33 | 14.22 | 67 % |
| 16 | 0.80 | 10.24 | 80 % |
Maximum Power vs Maximum Efficiency
At RL = Rth the load gets the most power, but only half of the Thevenin source's power reaches it (η = 50 %); the other half is lost in Rth. Efficiency keeps rising as RL grows, so power systems that need high efficiency use RL ≫ Rth, while signal and communication circuits (antennas, amplifiers) match impedances to get maximum power. Note the 50 % figure refers to the Thevenin equivalent; the real source network may waste power in other places too.
6. Reciprocity Theorem
Statement
In a linear, bilateral network with a single independent source and no dependent sources, if a voltage source in one branch produces a current in a second branch, then placing the same voltage source in the second branch produces the same current in the first branch. The ratio of response to excitation (a transfer resistance) is unchanged when source and response are interchanged.
Figure 6: Reciprocity in a T-network. Moving the source to the other port and the ammeter to the first gives the same current.
When Reciprocity Does Not Apply
It fails for networks with dependent sources, nonlinear or unilateral elements (diodes, transistors), and in general when there is more than one independent source. Also, the source and the response must be a voltage source and a current (or a current source and a voltage) so that the ratio is a transfer resistance or conductance.
7. Millman's Theorem
Statement
When several voltage sources, each in series with a resistance, are connected in parallel, they can be replaced by a single voltage source V in series with a single resistance Req, where V is the common voltage across the parallel combination.
Millman's Formulas
Req = 1 / (1/R₁ + 1/R₂ + … + 1/Rn) | In conductance form: V = Σ(Vk Gk) / Σ Gk | Use the sign of each Vk relative to the chosen reference polarity
Figure 7: Millman's theorem collapses the three branches into one 12 V source with 1 Ω in series. Branch currents then follow from Ik = (Vk − V)/Rk.
Branch Current Check
I₁ = (10 − 12)/2 = −1 A, I₂ = (20 − 12)/4 = +2 A, I₃ = (8 − 12)/4 = −1 A. Sum = 0, as KCL requires when no external load is connected. A negative sign means current flows into that source.
8. Tellegen's Theorem
Statement
For any lumped network, the sum of the instantaneous powers absorbed by all b branches is zero: Σ vk ik = 0 (k = 1 to b), with all branch voltages and currents taken in associated reference directions.
The Dutch engineer Bernard Tellegen proved this in 1952. It follows directly from KVL and KCL, so it is a statement of conservation of power: power delivered by sources equals power absorbed by the other elements. It is very general: it holds for linear and nonlinear, passive and active, and time-varying elements. A more general form says that voltages from one network and currents from any other network with the same topology also satisfy Σ vk îk = 0.
Tellegen's Theorem
Equivalently: total power supplied = total power absorbed
Example: Check Power Balance
A 10 V source drives 2 Ω and 3 Ω in series. Current I = 10/5 = 2 A. With associated reference directions, the source (current leaves + terminal) has v·i = −20 W (supplies 20 W).
| Branch | v (V) | i (A) | v · i (W) |
|---|---|---|---|
| 10 V source | 10 | −2 | −20 |
| 2 Ω resistor | 4 | 2 | +8 |
| 3 Ω resistor | 6 | 2 | +12 |
| Total | 0 |
9. Substitution Theorem
Statement
If the voltage v across and current i through any branch of a network (with a unique solution) are known, that branch can be replaced by an independent voltage source of value v, an independent current source of value i, or a resistor of value v/i, without changing any other voltage or current in the network.
Replace by Voltage Source
An 8 V source with the same polarity as the branch voltage.
Replace by Current Source
A 2 A source with the same direction as the branch current.
Replace by Resistor
R = v / i = 8 V / 2 A = 4 Ω (possible when i ≠ 0).
Example
In Figure 3 with RL = 4 Ω, the load branch carries 2 A with 8 V across it. Replacing it by an 8 V source, a 2 A source or a 4 Ω resistor leaves the rest of the circuit unchanged. The theorem works for linear and nonlinear networks, and it is mainly a tool for simplifying and for proving other results.
10. Compensation Theorem
Statement
In a linear network, if the resistance R of a branch carrying current I is changed by ΔR, the change in every branch current is the same as that produced by a compensating voltage source of value ΔR · I placed in series with the changed branch (opposing the original current), with all other independent sources replaced by their internal resistances.
Compensation Theorem Formula
Rth = Thevenin resistance seen by the branch (without R) | New current I′ = I + ΔI
Example
In Figure 3 the load R = 4 Ω carries I = 2 A (Vth = 16 V, Rth = 4 Ω). Increase R to 6 Ω (ΔR = 2 Ω). Then ΔI = −2 × 2 / (4 + 4 + 2) = −0.4 A, so I′ = 1.6 A. Direct check: 16/(4 + 6) = 1.6 A ✓. The compensation theorem is used in sensitivity analysis and bridge-circuit unbalance calculations.
11. Comparison Table and Network Theorem Formulas
| Theorem | Applies To | Key Idea | Key Formula |
|---|---|---|---|
| Superposition | Linear networks (any number of independent sources) | Add responses of each source acting alone | I = I′ + I″ + … |
| Thevenin | Linear two-terminal networks (load may be nonlinear) | Voltage source + series resistance | IL = Vth/(Rth + RL) |
| Norton | Linear two-terminal networks | Current source + parallel resistance | IN = Vth/Rth; RN = Rth |
| Source transformation | Sources with a series or parallel resistance | Convert V + R ⇄ I ∥ R | Is = Vs/Rs |
| Maximum power transfer | Load fed by a linear source network | Match load to source resistance | RL = Rth; Pmax = Vth²/(4Rth) |
| Reciprocity | Linear, bilateral, single source, no dependent sources | Interchange source and response | V/I (transfer resistance) unchanged |
| Millman | Parallel branches of source + resistance | One equivalent source for all branches | V = ΣVkGk / ΣGk |
| Tellegen | Any lumped network (even nonlinear, time-varying) | Conservation of power | Σ vk ik = 0 |
| Substitution | Any network with a unique solution | Replace a branch by an equal source or resistor | R = v / i |
| Compensation | Linear networks | Resistance change = series voltage source | ΔI = −IΔR/(Rth + R + ΔR) |
Deactivating Sources — Quick Rule
| Source Type | Deactivated By | Used In |
|---|---|---|
| Independent voltage source | Short circuit (0 V) | Superposition, Rth, RN |
| Independent current source | Open circuit (0 A) | Superposition, Rth, RN |
| Dependent source | Never deactivated | Keep active; use a test source for Rth |
12. Thevenin–Norton–Maximum Power Calculator
Enter the Thevenin equivalent of your circuit and a load value to get the Norton current, the load current, voltage and power, the maximum-power condition and the efficiency. Defaults match the example in Figure 3.
Thevenin–Norton–Max Power Calculator
All values in volts, ohms. Press Calculate to update.
13. Network Theorems Solved Problems (Numericals)
These network theorems numericals follow the standard exam pattern. Work each one yourself before reading the solution.
A 24 V source feeds a 6 Ω series resistor, followed by a 12 Ω resistor connected across terminals a–b. Find the Thevenin equivalent and the current in a 4 Ω load connected across a–b.
SolutionVth = 24 × 12/(6 + 12) = 16 V. Rth = 6 ∥ 12 = 72/18 = 4 Ω. IL = 16/(4 + 4) = 2 A.
Find the Norton equivalent of the circuit in Problem 1 and the current in a 12 Ω load.
SolutionShort a–b: IN = 24/6 = 4 A. RN = Rth = 4 Ω. IL = 4 × 4/(4 + 12) = 1 A. (Check with Thevenin: 16/(4 + 12) = 1 A ✓.)
In Figure 2, a 12 V source with 2 Ω feeds node A, which has a 4 Ω resistor to ground and a 6 A current source. Find VA.
Solution12 V alone (6 A open): VA′ = 12 × 4/(2 + 4) = 8 V. 6 A alone (12 V shorted): VA″ = 6 × (2 ∥ 4) = 6 × 1.333 = 8 V. VA = 8 + 8 = 16 V.
A network has Vth = 20 V and Rth = 10 Ω. Find RL for maximum power, the maximum power, and the power if RL = 20 Ω.
SolutionRL = Rth = 10 Ω. Pmax = 20²/(4 × 10) = 400/40 = 10 W. For RL = 20 Ω: I = 20/30 = 0.667 A, P = 0.667² × 20 = 8.89 W (less than Pmax ✓).
Two branches are in parallel: 10 V with 5 Ω, and 20 V with 5 Ω (both sources with + on the same node). Find the common voltage.
SolutionV = (10/5 + 20/5) / (1/5 + 1/5) = (2 + 4)/0.4 = 15 V; Req = 5 ∥ 5 = 2.5 Ω.
Convert a 20 V source in series with 5 Ω to its Norton form, then find the current in a 15 Ω load.
SolutionIs = 20/5 = 4 A in parallel with 5 Ω. Load current (current divider) = 4 × 5/(5 + 15) = 1 A. Check: 20/(5 + 15) = 1 A ✓.
A branch of 4 Ω carries 2 A. The Thevenin resistance seen by the branch is 4 Ω. If the branch resistance rises to 6 Ω, find the change in current.
SolutionΔI = −I ΔR / (Rth + R + ΔR) = −2 × 2/(4 + 4 + 2) = −0.4 A (current falls to 1.6 A).
Practice Questions (Answers in the Dropdown)
- A network has Vth = 12 V and Rth = 3 Ω. Find Pmax.
Show answer
RL = 3 Ω; Pmax = 144/(4 × 3) = 12 W. - A Norton equivalent has IN = 5 A, RN = 2 Ω. Find the Thevenin equivalent.
Show answer
Vth = 5 × 2 = 10 V, Rth = 2 Ω. - A voltage source of 30 V with 6 Ω series resistance is connected to a 4 Ω load. Find the load current by Thevenin and by Norton.
Show answer
Thevenin: 30/(6 + 4) = 3 A. Norton: IN = 5 A, load current = 5 × 6/(6 + 4) = 3 A.
14. Network Theorems MCQ (with Answers)
These network theory MCQs cover the questions most often asked in university exams and competitive tests such as GATE. Click "Show answer" to check.
- Voltage
- Current
- Power
- Both voltage and current
Show answer
C — Power. Power is proportional to the square of voltage or current, so it is not linear and cannot be superposed.
- An open circuit
- A short circuit
- A resistor
- A capacitor
Show answer
B — A short circuit (0 V). A deactivated current source becomes an open circuit.
- Short-circuit voltage
- Open-circuit voltage across the terminals
- Source voltage
- Load voltage
Show answer
B — The open-circuit voltage across the load terminals with the load removed.
- Open-circuit current
- Short-circuit current between the terminals
- Load current
- Total source current
Show answer
B — The short-circuit current through the terminals; IN = Vth/Rth.
- Zero
- Infinity
- Equal to the Thevenin resistance
- Twice the Thevenin resistance
Show answer
C — Equal to Rth.
- 25 %
- 50 %
- 75 %
- 100 %
Show answer
B — 50 %. Half of the power is dissipated in Rth and half in the load.
- 2.5 W
- 5 W
- 10 W
- 20 W
Show answer
B — 5 W. Pmax = Vth²/(4Rth) = 100/20 = 5 W.
- Nonlinear networks
- Linear bilateral networks with a single source
- Networks with dependent sources
- Unilateral networks
Show answer
B — Linear bilateral networks with a single independent source and no dependent sources.
- Power in a load
- The common voltage across parallel branches containing sources
- The short-circuit current
- The time constant
Show answer
B — The common voltage across parallel branches, V = ΣVkGk/ΣGk.
- Conservation of charge only
- Conservation of power (KVL and KCL)
- Linearity
- Bilateral property
Show answer
B — Conservation of power, derived from KVL and KCL. It does not require linearity.
- RN = 1/Rth
- RN = Rth
- RN = 2Rth
- RN = Rth/2
Show answer
B — RN = Rth.
- Compensation
- Reciprocity
- Substitution
- Millman
Show answer
C — Substitution theorem.
15. Network Theory Notes: Quick Revision
One-Page Revision Sheet
- Superposition: linear only; add partial responses; short voltage sources, open current sources; not for power
- Thevenin: Vth = open-circuit voltage; Rth = resistance with independent sources off; IL = Vth/(Rth + RL)
- Norton: IN = short-circuit current; RN = Rth; IN = Vth/Rth
- Source transformation: Vs + Rs ⇄ Is = Vs/Rs ∥ Rs
- Max power: RL = Rth; Pmax = Vth²/(4Rth); η = 50 %; AC: ZL = Zth*
- Reciprocity: linear, bilateral, single source; swap source and response, ratio is the same
- Millman: V = ΣVkGk/ΣGk; Req = 1/ΣGk
- Tellegen: Σ vk ik = 0 for any lumped network
- Substitution: branch (v, i) → source v, source i, or resistor v/i
- Compensation: ΔR in a branch ≡ series source ΔR · I; ΔI = −IΔR/(Rth + R + ΔR)
- Dependent sources: never deactivated; use a test source for Rth
Common Mistakes to Avoid
Exam Pitfalls
- Forgetting to remove the load before finding Vth or Rth.
- Shorting a current source or opening a voltage source when deactivating — it is the opposite.
- Deactivating a dependent source.
- Applying superposition directly to power.
- Getting the sign wrong when adding partial responses in superposition, or using the wrong polarity in source transformation.
- Using RL = Rth for AC circuits with reactance; the condition is ZL = Zth* (complex conjugate).
Frequently Asked Questions (FAQ)
Network theorems are rules in electrical network theory that simplify the analysis of complex circuits. Instead of solving every loop and node equation, a network theorem lets you replace a part of the circuit with a simpler equivalent or find one quantity quickly. The main network theorems are superposition, Thevenin's, Norton's, maximum power transfer, reciprocity, Millman's, Tellegen's, substitution and compensation theorems.
Thevenin's theorem states that any linear two-terminal network containing sources and resistances can be replaced by a single voltage source Vth in series with a single resistance Rth. Vth is the open-circuit voltage across the two terminals, and Rth is the resistance seen from the terminals with all independent sources deactivated (voltage sources shorted, current sources opened).
Norton's theorem states that any linear two-terminal network can be replaced by a single current source IN in parallel with a single resistance RN. IN is the short-circuit current between the terminals, and RN equals the Thevenin resistance Rth. A Norton equivalent and a Thevenin equivalent are related by source transformation: IN = Vth/Rth.
The superposition theorem states that in a linear network with more than one independent source, the voltage across or current through any element equals the algebraic sum of the voltages or currents produced by each independent source acting alone, with all other independent sources deactivated (voltage sources replaced by short circuits and current sources by open circuits). It works for voltage and current but not for power, because power is not a linear function.
The maximum power transfer theorem states that a source delivers maximum power to a load when the load resistance equals the Thevenin (source) resistance of the network feeding it, RL = Rth. The maximum power is Pmax = Vth²/(4Rth). At this condition the efficiency is 50 percent. For AC circuits the condition is that the load impedance equals the complex conjugate of the source impedance, ZL = Zth*.
The reciprocity theorem states that in a linear, bilateral network containing a single independent source and no dependent sources, if a source in one branch produces a current in a second branch, then moving the same source to the second branch produces the same current in the first branch. Equivalently, the ratio of response to excitation is unchanged when the source and the measuring point are interchanged.
Millman's theorem gives a single equivalent voltage source for several voltage sources, each in series with a resistance, connected in parallel. The common voltage is V = (V₁/R₁ + V₂/R₂ + … + Vn/Rn) / (1/R₁ + 1/R₂ + … + 1/Rn), and the equivalent series resistance is the parallel combination of all the resistances.
Tellegen's theorem states that for any lumped network, the sum of the instantaneous powers absorbed by all branches is zero: Σ vk ik = 0, using associated reference directions. It follows from Kirchhoff's voltage and current laws, so it holds for linear or nonlinear, passive or active, and time-varying elements, and it expresses conservation of power.
Thevenin's theorem replaces a network with a voltage source in series with a resistance, while Norton's theorem replaces it with a current source in parallel with a resistance. The resistances are equal (Rth = RN), and the sources are related by Vth = IN × Rth. Both give the same behaviour at the two terminals, so you can convert between them using source transformation.
First check the conditions (linear, bilateral, single or multiple sources). Then follow the standard steps of the chosen theorem: for Thevenin, remove the load, find the open-circuit voltage, deactivate independent sources to find Rth, redraw the equivalent and reconnect the load. For superposition, take one independent source at a time and add the results with signs. Always keep units consistent and verify the answer with a quick nodal or mesh check.