Thevenin Equivalent — Worked GATE EE Example

Prepared for GATE EE Study Hub · Original Electrical Engineering concept practice · 21 October 2026

Why Thevenin’s theorem matters

Many GATE EE circuit questions become simpler once a linear, two-terminal network is replaced by a single equivalent voltage source and a series resistance. This reduces repeated calculations when the load changes.

Example circuit

A 12 V ideal DC source is connected in series with a 4 Ω resistor, followed by an output node. An 8 Ω resistor connects that output node to ground. A load resistor R_L is connected between the output node and ground.

Find the open-circuit voltage V_th.
Find the Thevenin resistance R_th.
Find load current and voltage when R_L = 4 Ω.
Find the load resistance and power for maximum power transfer.

Step 1: Remove the load and calculate V_th

Without the load, the circuit is a voltage divider formed by 4 Ω and 8 Ω. Therefore:

V_th = 12 × 8 / (4 + 8) = 8 V

The Thevenin voltage is 8 V at the output relative to ground.

Step 2: Find R_th

Deactivate the independent ideal 12 V voltage source by replacing it with a short circuit. Looking into the output terminals, the 4 Ω and 8 Ω resistors are now in parallel:

R_th = (4 × 8) / (4 + 8) = 8 / 3 Ω ≈ 2.6667 Ω

This rule applies to independent ideal voltage sources. If a circuit contains dependent sources, retain them and use an appropriate test-source method to find R_th.

Step 3: Connect the 4 Ω load

The circuit is now equivalent to an 8 V source in series with 8/3 Ω and the 4 Ω load.

I_L = 8 / ((8/3) + 4) = 1.2 A

V_L = I_L × 4 = 4.8 V

Using the original circuit directly gives the same result: 8 Ω || 4 Ω = 8/3 Ω; a divider with the first 4 Ω resistor gives 4.8 V at the load node.

Step 4: Maximum power transfer

For a purely resistive network, load power is maximum when R_L = R_th.

R_L,opt = R_th = 8 / 3 Ω

P_max = V_th² / (4 R_th) = 8² / (4 × 8/3) = 6 W

Answer	Value
Open-circuit Thevenin voltage	8 V
Thevenin resistance	2.6667 Ω
Current into 4 Ω load	1.2 A
Voltage across 4 Ω load	4.8 V
Load for maximum power	2.6667 Ω
Maximum load power	6 W

Common pitfalls

Removing the wrong resistor when calculating the open-circuit terminal voltage.
Short-circuiting a current source instead of opening it when deactivating independent sources.
Deactivating dependent sources, which generally must remain active.
Applying maximum power transfer without checking what network and load assumptions are being used.

Self-check

For the same Thevenin equivalent, what is the load current when R_L = 8 Ω?

Show answer

I_L = 8 / (8/3 + 8) = 0.75 A; the load voltage is 0.75 × 8 = 6 V.

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Thevenin’s theorem, without guesswork.