Superposition, Thevenin and Norton theorems

Why it matters

Most instrumentation problems ask what one load sees: a meter across a bridge, an ADC input on a sensor, a cable on a transducer. Thevenin and Norton theorems replace everything except that load with one source and one resistance, so loading errors and output swings can be read off immediately. Superposition lets you study each excitation separately, for example the signal and the bias or offset, and is the basis of every linear-system idea that follows.

Key ideas

Linearity. A circuit is linear if it contains only linear elements (R, L, C, linear dependent sources) and independent sources. For such circuits every response is a weighted sum of the independent sources, which gives two properties:

• Homogeneity: doubling all sources doubles every response.
• Additivity: the response to several sources is the sum of the responses to each.

Superposition theorem. In a linear circuit with several independent sources, any branch current or voltage is the algebraic sum of the responses produced by each independent source acting alone. To make a source "act alone", deactivate the others:

• An independent voltage source is replaced by a short circuit (V = 0).
• An independent current source is replaced by an open circuit (I = 0).
• Dependent sources are never deactivated. They stay in the circuit in every sub-problem.
• Practical sources keep their internal resistance.

Superposition does not apply to power, because power is quadratic in current or voltage: (I1 + I2)²R ≠ I1²R + I2²R. Find the total current first, then compute power. Superposition is also the only correct way to handle sources of different frequencies (for example DC plus AC): solve each frequency separately and add in the time domain.

Thevenin's theorem. Any linear two-terminal network can be replaced at its terminals by a voltage source Vth in series with a resistance Rth (an impedance Zth in AC).

• Vth is the open-circuit voltage at the terminals.
• Rth is the resistance seen into the terminals with all independent sources deactivated.

Norton's theorem. The same network can be replaced by a current source IN in parallel with RN.

• IN is the short-circuit current at the terminals.
• RN = Rth.

The two forms are related by source transformation: Vth = IN·Rth.

Three ways to find Rth.

1. Only independent sources present: deactivate them and reduce by series–parallel.
2. Any circuit: Rth = Voc/Isc.
3. Dependent sources present: deactivate the independent sources only. Apply a test source (Vt or It) at the terminals and use Rth = Vt/It.

With dependent sources, Rth can come out zero or even negative, which indicates an active circuit. If Vth = 0 and IN = 0 (no independent sources), only method 3 works.

Loading. A Thevenin source driving a load RL delivers VL = Vth·RL/(Rth + RL). An instrument with input resistance Rin reads Vth·Rin/(Rth + Rin). This is why voltmeters need Rin much larger than Rth.

Formulas

• Superposition: x = Σ x_k, where x_k is the response with source k acting alone.
• Thevenin voltage: Vth = Voc (V).
• Norton current: IN = Isc (A).
• Equivalent resistance: Rth = RN = Voc/Isc = Vt/It (Ω).
• Source conversion: Vth = IN·Rth.
• Load voltage: VL = Vth·RL/(Rth + RL). Load current: IL = Vth/(Rth + RL).
• Loading error of a meter: e = Rth/(Rth + Rin), as a fraction of Vth.

These hold for linear networks, in DC or with phasors and impedances in AC steady state.

Worked examples

Example 1 (standard): superposition. Given: a 20 V source in series with 4 Ω feeds node a. Node a is connected to ground through 6 Ω. A 2 A current source injects current into node a from ground. Find the current in the 6 Ω resistor.

1. 20 V source alone (2 A source opened): I' = 20/(4 + 6) = 2.0 A.
2. 2 A source alone (20 V source shorted): the 4 Ω and 6 Ω resistors are in parallel across the source. By current division, I'' = 2·4/(4 + 6) = 0.8 A, flowing down through 6 Ω.
3. Add: I = 2.0 + 0.8 = 2.8 A.
4. Check by nodal analysis: (V − 20)/4 + V/6 = 2, so V·(5/12) = 7, which gives V = 16.8 V and I = 16.8/6 = 2.8 A. ✓

Answer: 2.8 A.

Example 2 (GATE level): Thevenin equivalent with a dependent source. Given:

• A 12 V source in series with 3 Ω drives node a. Call the current in the 3 Ω resistor (toward a) i1.
• Node a connects to ground through 6 Ω.
• A current-controlled current source draws 0.5·i1 from node a to ground.
• The terminals are a and ground.

Find the Thevenin equivalent and the current into a 3 Ω load.

1. Open-circuit voltage. KCL at a: i1 = V/6 + 0.5·i1, so 0.5·i1 = V/6. With i1 = (12 − V)/3, this gives (12 − V)/6 = V/6, so Vth = 6 V.
2. Short-circuit current. With a shorted, V = 0, so i1 = 12/3 = 4 A. The 6 Ω resistor carries nothing, and the dependent source takes 0.5·4 = 2 A. That leaves Isc = 4 − 2 = 2 A.
3. Rth = Voc/Isc = 6/2 = 3 Ω.
4. Test-source check (12 V shorted, apply Vt at a): i1 = −Vt/3. The test current is It = Vt/6 + 0.5·i1 − i1 = Vt/6 + Vt/6 = Vt/3, so Rth = 3 Ω. ✓
5. With a 3 Ω load, IL = 6/(3 + 3) = 1 A.

Answer: Vth = 6 V, Rth = 3 Ω, IL = 1 A.

Common mistakes

• Deactivating dependent sources in superposition or when finding Rth. Only independent sources are set to zero.
• Opening a voltage source or shorting a current source when deactivating. It is the other way round.
• Adding powers from each source. Add currents or voltages, then compute power.
• Finding Rth by series–parallel reduction when dependent sources are present. Use Voc/Isc or a test source.
• Losing the polarity of Vth or the direction of IN. Keep a consistent "+ at terminal a" reference.
• Forgetting that a resistor directly across an ideal voltage source, or in series with an ideal current source, does not affect the Thevenin equivalent seen elsewhere.

For GATE IN

Common question types: find Vth and Rth of a resistive network containing a dependent source (often NAT); find a branch current by superposition with one voltage and one current source; find the reading of a voltmeter of given input resistance across a network (loading error); find a DC plus AC response by superposition. Practise the Voc/Isc and test-source methods until they are quick, and always check Rth two ways when time allows.

Quick check

1. When applying superposition, what replaces an independent current source?
2. Voc = 10 V and Isc = 2.5 A at a pair of terminals. What is Rth?
3. Can superposition give the total power in a resistor directly?
4. A 10 V, 1 kΩ Thevenin source is read by a 10 kΩ voltmeter. What does the meter show?

Answers: 1. An open circuit. 2. 4 Ω. 3. No; superpose currents, then square. 4. 10 × 10/11 = 9.09 V.