Network theorems: Thevenin, Norton and superposition
Superposition, Thevenin and Norton equivalents (including dependent sources and the test-source method) and maximum power transfer, with standard, GATE-level and superposition examples.
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Why it matters
A sensor bridge feeding an ADC, a battery feeding a motor driver, or an amplifier output driving a load are all "a network seen from two terminals". Network theorems let you replace everything behind those terminals with one source and one resistance, so you can try many loads without re-solving the whole circuit. They also explain loading errors and maximum power transfer, which matter whenever a sensor or source has internal resistance.
Key ideas
Linearity. The theorems here apply to linear, bilateral networks: resistors (or impedances), independent sources, and linear dependent sources. A circuit is linear if its response obeys homogeneity (double the source, double the response) and additivity. Diodes, saturating transistors and lamps with temperature-dependent resistance are not linear.
Superposition. In a linear circuit with several independent sources, any voltage or current equals the algebraic sum of the contributions from each independent source acting alone.
- To "switch off" a source: replace an ideal voltage source by a short circuit and an ideal current source by an open circuit. Keep any internal resistance in place.
- Dependent sources are never switched off; they stay in every sub-circuit.
- Power is not linear in current (P = I²R), so you superpose currents and voltages, then compute power from the total.
- Superposition is especially useful when sources are of different kinds or frequencies (for example a DC bias plus an AC signal).
Thevenin's theorem. Any linear two-terminal network can be replaced by a voltage source V_th in series with a resistance R_th.
- V_th is the open-circuit voltage at the terminals (load removed).
- R_th is the resistance seen into the terminals with all independent sources switched off.
- If the network contains dependent sources, either use R_th = V_oc / I_sc, or apply a test source V_t at the terminals (independent sources off) and compute R_th = V_t / I_t. A dependent source can even make R_th negative.
Norton's theorem. The same network can be replaced by a current source I_N in parallel with R_N, where I_N is the short-circuit current at the terminals and R_N = R_th. Thevenin and Norton forms are interchangeable by source transformation, except when R_th = 0 (no Norton form) or R_th → ∞ (no Thevenin form).
Maximum power transfer. A fixed source (V_th, R_th) delivers maximum power to a resistive load when R_L = R_th. The efficiency at that point is only 50 %, so the condition is used for signal circuits (sensors, RF, audio) and not for power systems, where R_L ≫ R_th is wanted. For AC, the load must be the complex conjugate of the source impedance.
Other theorems you will meet. Reciprocity (in a passive bilateral network, the ratio of response to excitation is unchanged if their positions are swapped), Millman's theorem (parallel voltage sources with series resistances combine into one), and the substitution theorem.
Formulas
V_th = V_oc (open-circuit terminal voltage)
- V_th, V_oc: volts (V).
I_N = I_sc (short-circuit terminal current)
- I_N, I_sc: amperes (A).
R_th = R_N = V_oc / I_sc
- R_th, R_N: Ω. Works with or without dependent sources.
R_th = V_t / I_t (test-source method, independent sources off)
- V_t: applied test voltage (V); I_t: current drawn from it into the network (A).
V_th = I_N · R_th (source transformation)
I_L = V_th / (R_th + R_L)
- I_L: load current (A); R_L: load resistance (Ω).
P_max = V_th² / (4·R_th) at R_L = R_th
- P_max: maximum power to the load (W). Valid for a resistive load and fixed source.
η = R_L / (R_th + R_L)
- η: fraction of the source power delivered to the load; equals 0.5 at maximum power transfer.
Worked examples
Example 1 (standard: Thevenin, Norton, maximum power). A 12 V source in series with R1 = 4 Ω feeds node X. R2 = 12 Ω connects X to ground, and R3 = 2 Ω connects X to terminal A; terminal B is ground. Find the Thevenin and Norton equivalents at A–B, the current in a 10 Ω load, and the maximum power any load can draw.
- With A–B open, no current flows in R3, so
V_th = 12 × R2/(R1 + R2) = 12 × 12/16 = 9 V. - Switch off the 12 V source (short).
R_th = (R1 ∥ R2) + R3 = (4 × 12)/(4 + 12) + 2 = 3 + 2 = 5 Ω. - Norton:
I_N = V_th/R_th = 9/5 = 1.8 A, in parallel with 5 Ω. - With R_L = 10 Ω:
I_L = 9/(5 + 10) = 0.6 A, so P_L = 0.6² × 10 = 3.6 W. - Maximum power at R_L = 5 Ω:
P_max = 9²/(4 × 5) = 81/20 = 4.05 W.
V_th = 9 V, R_th = 5 Ω, I_N = 1.8 A; I_L = 0.6 A in 10 Ω; P_max = 4.05 W at R_L = 5 Ω.
Example 2 (GATE level: Thevenin with a dependent source). A 10 V source in series with R1 = 2 Ω feeds terminal A; let I_x be the current in R1 flowing towards A. R2 = 4 Ω connects A to ground, and a dependent current source draws 0.5·I_x from A to ground. Terminal B is ground. Find the Thevenin equivalent at A–B.
- Open circuit: I_x = (10 − V)/2. KCL at A (currents leaving):
V/4 + 0.5·I_x − I_x = 0, so V/4 = 0.5·I_x = 0.25·(10 − V). - V = 10 − V, so
V_oc = 5 V. - Short circuit (V = 0): I_x = 10/2 = 5 A; the dependent source takes 0.5 × 5 = 2.5 A and R2 carries none, so
I_sc = 5 − 2.5 = 2.5 A. R_th = V_oc/I_sc = 5/2.5 = 2 Ω.- Test-source check: switch off the 10 V source and apply V_t at A. Then I_x = −V_t/2 and the test current is I_t = V_t/4 + 0.5·I_x − I_x = V_t/4 + V_t/4 = V_t/2, so R_th = 2 Ω again.
- Note that simply combining resistors would give 2 ∥ 4 = 1.33 Ω, which is wrong because the dependent source changes the terminal behaviour.
V_th = 5 V, R_th = 2 Ω (I_N = 2.5 A); the maximum load power is 5²/(4 × 2) = 3.125 W.
Example 3 (superposition and power). A 12 V source with 4 Ω feeds node A; 6 Ω connects A to ground; a 2 A current source also injects into A.
- 12 V alone (2 A source open): V_A' = 12 × 6/10 = 7.2 V.
- 2 A alone (12 V source shorted): V_A'' = 2 × (4 ∥ 6) = 2 × 2.4 = 4.8 V.
- Total V_A = 12.0 V, so the 6 Ω resistor carries 2 A and absorbs 24 W, while adding the individual powers (8.64 W + 3.84 W = 12.48 W) is wrong.
V_A = 12 V; P(6 Ω) = 24 W.
Common mistakes
- Switching off a dependent source when finding R_th or applying superposition.
- Shorting a current source or opening a voltage source when switching sources off (it is the other way round).
- Leaving the load connected while finding V_oc.
- Superposing powers instead of currents or voltages.
- Believing maximum power transfer means maximum efficiency; efficiency is only 50 % there.
- Using R_th = V_oc/I_sc when both are zero (a network with only dependent sources); use a test source instead.
For GATE ME
Typical questions ask for V_th, R_th or I_N at marked terminals of a two- or three-loop network, the load that receives maximum power and that power, a branch current by superposition with one voltage and one current source, or R_th of a circuit containing a dependent source. Practise the test-source method and quick series-parallel reduction, and check each answer with V_th = I_N·R_th.
Quick check
- When switching off sources for superposition, what replaces an ideal current source?
- A network has V_oc = 20 V and I_sc = 4 A. What is R_th?
- For the network in question 2, what is the maximum power deliverable to a load?
- Why can't you find R_th by series-parallel reduction when a dependent source is present?
Answers: 1. An open circuit. 2. 5 Ω. 3. 20²/(4 × 5) = 20 W. 4. Its output depends on circuit variables, so it must stay active; use V_oc/I_sc or a test source.
Interview questions
All Electrical Circuits and Electronics interview questionsTry answering each one aloud before you open it.
1.What is Thevenin's theorem and how is it used in circuit analysis?Concept
Thevenin's theorem says any linear two-terminal network can be replaced by a voltage source V_th in series with a resistance R_th. V_th is the open-circuit voltage at the terminals, and R_th is the resistance seen into the terminals with independent sources switched off (or V_oc/I_sc if dependent sources are present). It lets you analyse different loads, loading errors and maximum power transfer without re-solving the whole network each time.
2.Explain Norton's theorem and its application in electrical circuits.Concept
Norton's theorem replaces a linear two-terminal network with a current source I_N in parallel with a resistance R_N. I_N is the short-circuit current at the terminals and R_N equals the Thevenin resistance, so I_N = V_th/R_th. The Norton form is convenient when loads are connected in parallel or when the network is naturally described by current sources, such as transistor output stages or current-output sensors.
3.What is the superposition theorem and when is it applicable?Concept
The superposition theorem states that in a linear circuit with multiple independent sources, the voltage across or current through any element is the algebraic sum of the voltages or currents produced by each source acting independently. It is applicable in linear circuits where the principle of linearity holds, meaning the circuit's response is directly proportional to the input.
4.Why is Thevenin's theorem preferred over Norton's theorem in certain applications?Application
Thevenin's theorem is often preferred when dealing with voltage sources and series connections, as it simplifies the circuit to a single voltage source and series resistance. This can make it easier to analyze the impact of connecting additional series components. Conversely, Norton's theorem is more convenient for circuits with current sources and parallel connections.
5.How does the superposition theorem help in analyzing circuits with multiple sources?Application
Superposition lets you solve a linear circuit one independent source at a time and add the results. Each other independent voltage source is replaced by a short and each current source by an open, while dependent sources stay active. It is most useful when sources differ in kind or frequency, such as a DC bias plus an AC signal. You superpose voltages and currents, never powers.
6.What happens if you apply Thevenin's theorem to a non-linear circuit?Application
Thevenin's theorem is not applicable to non-linear circuits because it relies on the principle of linearity. In non-linear circuits, the relationship between voltage and current is not proportional, so the equivalent circuit representation provided by Thevenin's theorem would not accurately reflect the circuit's behavior.
7.Can Norton's and Thevenin's theorems be used interchangeably? Why or why not?Application
Usually yes: the two equivalents are related by source transformation, V_th = I_N·R_th with the same resistance, so either gives identical terminal behaviour. The exceptions are the limiting cases: a network with R_th = 0 (an ideal voltage source) has no Norton equivalent, and one with infinite R_th (an ideal current source) has no Thevenin equivalent. Both describe only the terminal behaviour, not the internal power dissipation of the original network.
8.A 10 V source is in series with 4 Ω and 6 Ω resistors, and the output terminals are taken across the open ends of this series string (source and both resistors in the path). Find the Thevenin equivalent.Numerical
With the terminals open, no current flows, so there is no drop across either resistor and V_th = 10 V. Switching off the source (short circuit) leaves the two resistors in series between the terminals, so R_th = 4 + 6 = 10 Ω. The equivalent is 10 V in series with 10 Ω.
9.A 5 A current source is in parallel with 3 Ω and 2 Ω resistors, and the output terminals are across this parallel combination. Find the Norton equivalent.Numerical
Shorting the terminals also shorts both resistors, so all 5 A flows through the short: I_N = 5 A. With the current source switched off (open), the resistance seen is 3 ∥ 2 = (3 × 2)/(3 + 2) = 1.2 Ω. The Norton equivalent is 5 A in parallel with 1.2 Ω (equivalently 6 V in series with 1.2 Ω).
10.What are the limitations of using the superposition theorem in circuit analysis?Application
It applies only to linear circuits, so it cannot be used directly with diodes, saturating transistors or other non-linear elements (only with their small-signal linear models). It cannot be applied to power, because power depends on the square of current or voltage; you must add the currents first and then compute power. Dependent sources cannot be switched off, and with many sources the repeated sub-circuits can take longer than a single nodal analysis.
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