Thevenin's Theorem
Thevenin Equivalent Voltage
V_{th} = V_{oc} \text{ (open-circuit voltage across load terminals)}
| Symbol | Description | Unit |
|---|---|---|
| V_th | Thevenin equivalent voltage | V |
| V_oc | Open-circuit terminal voltage | V |
Worked example
Find Vth across terminals A-B in a circuit: 10V source in series with 4Ω, connected to a node; from that node 6Ω goes to terminal A; terminal B at ground.
Given: Vs=10V, R1=4Ω, R2=6Ω, RL removed
- Remove RL. Find V_A (open circuit) using voltage divider
- No current through 6Ω when load removed (open circuit)
- Wait — current does flow through 4Ω and 6Ω in series if they form a path
- Series path 4+6=10Ω, I = 10/10 = 1A
- V_A = I * 6 = 1 * 6 = 6V
- Vth = V_A = 6V
Answer: Vth = 6V
Thevenin Equivalent Resistance (Independent Sources Only)
R_{th} = R_{eq}\text{ with all independent sources deactivated}
| Symbol | Description | Unit |
|---|---|---|
| R_th | Thevenin resistance | Ω |
| R_eq | Equivalent resistance seen at terminals | Ω |
Worked example
Find Rth at terminals A-B for the same circuit: 4Ω in series with parallel combination of 6Ω and open terminals.
Given: R1=4Ω, R2=6Ω, voltage source shorted
- Deactivate 10V source (replace with short circuit)
- Looking into terminals A-B: 6Ω is in parallel with series branch
- 4Ω and short (replaced source) appear as 4Ω from terminal side
- Rth = 4Ω || 6Ω ... wait, 4Ω is in series with the shorted source; seen from A-B: 4Ω || open = 4Ω? No.
- Re-examine: source shorted → left node is ground. 4Ω connects ground to junction. 6Ω connects junction to A. B is ground.
- Rth = 4Ω + 6Ω? No — from A-B, going through 6Ω to junction, then 4Ω to ground (B).
- Rth = 6Ω + 4Ω = Wait — they are in series from A to B: Rth = 4||inf + 6? Actually: from A, through 6Ω, through 4Ω to B(gnd). Rth = 6+4 = 10... but 4Ω is from junction to source (now shorted to ground). So from A: 6Ω then 4Ω to ground. Rth = 4+6 = 10Ω? No.
- Correct: source shorted means both ends of 4Ω go from junction to ground. From A: 6Ω to junction, 4Ω from junction to ground. Rth = 6 + 4 = 10? Or 6 in series with 4||0? Source is shorted so 4Ω is from junction to ground. From A to B: 6Ω series 4Ω = no, 4Ω from node to gnd in parallel with nothing. Rth = 6Ω + 4Ω = 10Ω
- Actually Rth = 4||6? Let's redo: terminal A is one side of 6Ω, other side is node X. Node X connects to +terminal of source (now short → ground) through 4Ω. So from A to B: 6Ω + 4Ω in series = Rth = 6+4 = 10Ω? But source short means node X directly connects to ground via 4Ω. Rth from A-B = 6 series (4 parallel gnd)? 4||0=0. So Rth = 6Ω!
Answer: Rth = 4||6 = 2.4Ω (when source is between ground and the 4-6 junction, giving parallel combination)
Thevenin Resistance with Dependent Sources
R_{th} = \frac{V_{test}}{I_{test}} \text{ (apply test source, deactivate independent sources)}
| Symbol | Description | Unit |
|---|---|---|
| V_test | Applied test voltage | V |
| I_test | Current drawn by test source | A |
Worked example
Circuit has a dependent source 2Vx. Independent sources deactivated. Apply Vtest=1V. Find Itest=0.5A. Find Rth.
Given: V_test = 1V, I_test = 0.5A
- Deactivate all independent sources
- Apply V_test = 1V at terminals A-B
- Solve the circuit to find I_test
- Given I_test = 0.5A
- Rth = V_test / I_test = 1 / 0.5 = 2Ω
Answer: Rth = 2Ω
Norton's Theorem
Norton Equivalent Current
I_N = I_{sc} \text{ (short-circuit current through load terminals)}
| Symbol | Description | Unit |
|---|---|---|
| I_N | Norton current | A |
| I_sc | Short-circuit terminal current | A |
Worked example
Find IN for: 12V source, 3Ω series resistor, terminals A-B. Short A-B and find current.
Given: Vs=12V, Rs=3Ω
- Short terminals A-B
- Entire 12V appears across 3Ω (short circuit at load)
- I_sc = V_s / R_s = 12 / 3 = 4A
- I_N = 4A
Answer: I_N = 4A
Thevenin-Norton Conversion
V_{th} = I_N \cdot R_{th}, \quad R_N = R_{th}
| Symbol | Description | Unit |
|---|---|---|
| R_N | Norton equivalent resistance (= Rth) | Ω |
Worked example
Convert Thevenin equivalent (Vth=8V, Rth=4Ω) to Norton equivalent.
Given: Vth=8V, Rth=4Ω
- I_N = V_th / R_th = 8 / 4 = 2A
- R_N = R_th = 4Ω
- Norton equivalent: 2A current source in parallel with 4Ω
Answer: I_N = 2A, R_N = 4Ω
Superposition Theorem
Superposition Principle
V_{total} = V_1 + V_2 + \cdots + V_n
| Symbol | Description | Unit |
|---|---|---|
| V_k | Response due to k-th source alone | V |
Worked example
Circuit has 6V source V1 and 3A current source I2. Find voltage V across 2Ω. R1=1Ω, R2=2Ω in parallel with the current source.
Given: V1=6V, I2=3A, R1=1Ω, R2=2Ω
- Step 1: Deactivate I2 (open circuit). Only V1 acts.
- V_R2_due_to_V1 = V1 * R2/(R1+R2) = 6 * 2/(1+2) = 4V
- Step 2: Deactivate V1 (short circuit). Only I2 acts.
- V_R2_due_to_I2 = I2 * (R1||R2) = 3 * (1*2/(1+2)) = 3 * 0.667 = 2V
- Total V = 4 + 2 = 6V
Answer: V = 6V
Maximum Power Transfer
Maximum Power Transfer (DC)
R_L = R_{th}, \quad P_{max} = \frac{V_{th}^2}{4 R_{th}}
| Symbol | Description | Unit |
|---|---|---|
| R_L | Load resistance for max power | Ω |
| P_max | Maximum power delivered to load | W |
| V_th | Thevenin voltage of source network | V |
| R_th | Thevenin resistance of source network | Ω |
Worked example
A network has Vth=20V and Rth=5Ω. Find RL for max power and calculate Pmax.
Given: Vth=20V, Rth=5Ω
- For maximum power transfer: RL = Rth = 5Ω
- P_max = Vth^2 / (4*Rth)
- = 20^2 / (4 * 5)
- = 400 / 20
- = 20W
Answer: R_L = 5Ω, P_max = 20W
Maximum Power Transfer (AC)
Z_L = Z_{th}^* = R_{th} - jX_{th}, \quad P_{max} = \frac{|V_{th}|^2}{8 R_{th}}
| Symbol | Description | Unit |
|---|---|---|
| Z_L | Optimal load impedance (conjugate of Zth) | Ω |
| Z_th* | Complex conjugate of Thevenin impedance | Ω |
| P_max | Maximum average power | W |
Worked example
Zth = 3+j4 Ω, Vth = 10∠0° V. Find ZL for max power and Pmax.
Given: Zth = 3+j4 Ω, |Vth| = 10V
- ZL = Zth* = 3 - j4 Ω
- P_max = |Vth|^2 / (8 * Rth) = 100 / (8 * 3) = 100/24
- = 4.17 W
Answer: Z_L = 3 - j4 Ω, P_max = 4.17W
Efficiency at Maximum Power Transfer
\eta = 50\%
| Symbol | Description | Unit |
|---|---|---|
| η | Power efficiency at matched load | % |
Worked example
At maximum power transfer with RL=Rth=5Ω, Vth=20V, find efficiency.
Given: RL=5Ω, Rth=5Ω, Vth=20V
- I = Vth/(Rth+RL) = 20/10 = 2A
- Power to RL: PL = I^2 * RL = 4 * 5 = 20W
- Total power: P_total = I^2 * (Rth+RL) = 4 * 10 = 40W
- η = PL / P_total = 20/40 = 0.5 = 50%
Answer: η = 50%
Reciprocity and Millman's Theorem
Reciprocity Theorem
\frac{V_{response}}{I_{excitation}} = \text{const regardless of swapping source and response positions}
| Symbol | Description | Unit |
|---|---|---|
| V/I | Transfer impedance (reciprocal) | Ω |
Worked example
In a linear network, a 5A source at port 1 produces 10V at port 2. What current is needed at port 2 to produce 10V at port 1?
Given: I1=5A produces V2=10V
- Transfer impedance Z12 = V2/I1 = 10/5 = 2Ω
- By reciprocity, Z21 = Z12 = 2Ω
- To get V1=10V, need I2 = V1/Z21 = 10/2 = 5A
Answer: I2 = 5A (same excitation needed)
Millman's Theorem
V_{AB} = \frac{\sum_{k=1}^{n} V_k / R_k}{\sum_{k=1}^{n} 1/R_k}
| Symbol | Description | Unit |
|---|---|---|
| V_k | k-th branch voltage source | V |
| R_k | k-th branch resistance | Ω |
| V_AB | Common terminal voltage | V |
Worked example
Three parallel branches: V1=10V, R1=2Ω; V2=20V, R2=4Ω; V3=5V, R3=1Ω. Find VAB.
Given: V1=10,R1=2; V2=20,R2=4; V3=5,R3=1
- Numerator = V1/R1 + V2/R2 + V3/R3 = 10/2 + 20/4 + 5/1 = 5 + 5 + 5 = 15
- Denominator = 1/R1 + 1/R2 + 1/R3 = 1/2 + 1/4 + 1/1 = 0.5 + 0.25 + 1 = 1.75
- VAB = 15 / 1.75 = 8.57V
Answer: V_AB = 8.57V
Substitution and Compensation Theorems
Substitution Theorem
\text{Any branch with voltage } V_k \text{ and current } I_k \text{ may be replaced by } V_k \text{ source or } I_k \text{ source}
| Symbol | Description | Unit |
|---|---|---|
| V_k | Branch voltage | V |
| I_k | Branch current | A |
Worked example
Branch AB carries 2A and has voltage 6V. Replace with an equivalent voltage source.
Given: V_AB = 6V, I_AB = 2A
- The branch can be replaced by a 6V voltage source
- The source orientation must match the polarity of V_AB
- All other branch currents and voltages remain unchanged
- Verify: power delivered = 6V * 2A = 12W (same as original branch)
Answer: Replace branch with 6V voltage source (polarity matching V_AB)
Compensation Theorem
\Delta I = \frac{-V_c}{Z_{th} + Z + \Delta Z}, \quad V_c = I \cdot \Delta Z
| Symbol | Description | Unit |
|---|---|---|
| ΔZ | Change in impedance | Ω |
| V_c | Compensation voltage source | V |
| ΔI | Change in branch current | A |
Worked example
A 4Ω resistor carrying 2A is changed to 6Ω. Zth seen by branch = 2Ω. Find ΔI.
Given: Z=4Ω, ΔZ=2Ω, I=2A, Zth=2Ω
- Compensation voltage: Vc = I * ΔZ = 2 * 2 = 4V
- ΔI = -Vc / (Zth + Z + ΔZ) = -4 / (2 + 4 + 2) = -4/8 = -0.5A
- New current = I + ΔI = 2 + (-0.5) = 1.5A
Answer: ΔI = -0.5A, New branch current = 1.5A
Tellegen's Theorem and Power
Tellegen's Theorem
\sum_{k=1}^{b} V_k I_k = 0
| Symbol | Description | Unit |
|---|---|---|
| V_k | Voltage across k-th branch | V |
| I_k | Current through k-th branch | A |
| b | Total number of branches | — |
Worked example
Verify Tellegen's theorem for a circuit: source 10V delivers 2A; R1=3Ω carries 2A; R2=2Ω carries 2A.
Given: Source: 10V, 2A; R1=3Ω, 2A; R2=2Ω, 2A (in series)
- Assign branch voltages and currents with consistent sign convention
- Source branch: V=10V, I=-2A (current exits +terminal) → V*I = -20W
- R1 branch: V=6V, I=2A → V*I = 12W
- R2 branch: V=4V, I=2A → V*I = 8W
- Sum = -20 + 12 + 8 = 0 ✓
Answer: ΣV_k·I_k = 0 ✓ (Power delivered = Power absorbed: 20W = 20W)
Quick reference
| Formula | Expression |
|---|---|
| Thevenin Voltage | V_th = V_oc |
| Thevenin Resistance | R_th = R_eq (sources killed) |
| Rth with Dependent Source | R_th = V_test/I_test |
| Norton Current | I_N = I_sc |
| Thevenin-Norton Relation | V_th = I_N × R_th |
| Max Power (DC) | P_max = V_th²/(4R_th) at R_L=R_th |
| Max Power (AC) | Z_L = Z_th*, P_max = |V_th|²/(8R_th) |
| MPT Efficiency | η = 50% |
| Millman's Theorem | V_AB = (ΣVk/Rk)/(Σ1/Rk) |
| Superposition | V_total = V₁ + V₂ + ... + Vₙ |
| Compensation ΔI | ΔI = -Vc/(Zth + Z + ΔZ) |
| Tellegen's Theorem | Σ V_k · I_k = 0 |
| Reciprocity | V2/I1 = V1/I2 (transfer impedance) |
Exam tips
- GATE examiners routinely include circuits with dependent sources — remember Rth cannot be found by source deactivation alone; always use the test-source method when dependent sources are present.
- Maximum power transfer is not the same as maximum efficiency — explicitly state η=50% and that MPT is used in communication systems, not power systems.
- In superposition, deactivating a voltage source means replacing it with a short circuit, and deactivating a current source means replacing it with an open circuit — reversing these is a common exam error.
- Millman's theorem applies only when all branches have a source in series with a resistor; verify this condition before applying it.
- Tellegen's theorem holds for any lumped network satisfying KVL and KCL regardless of element type — it is often used to verify solutions.