--- title: "Thevenin current through RL and maximum power" subject: "Fundamental of Electrical and Electronics Engineering" chapter: 1 key: "semester_1/fundamental_of_electrical_and_electronics_engineering:thevenin-max-power-load" source: https://www.bctengineer.com/q/semester_1/fundamental_of_electrical_and_electronics_engineering/thevenin-max-power-load site: BCT Engineering --- # Thevenin current through RL and maximum power Fundamental of Electrical and Electronics Engineering · Chapter 1: Basic Circuits Concepts - Asked in: 2081 Chaitra (new course), 6+2 marks - Video: [Maximum Power Transfer Theorem Using Nodal Analysis & Thevenin Equivalent Circuits](https://www.youtube.com/watch?v=8CA6ZNXgI-Y) — The Organic Chemistry Tutor - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/fundamental_of_electrical_and_electronics_engineering/thevenin-max-power-load - Whole chapter: https://www.bctengineer.com/md/semester_1/fundamental_of_electrical_and_electronics_engineering/chapter-1 ## Question In a given network, find the current flowing through the load resistance RL by using Thevenin's theorem. What should be the value of RL for maximum power to transfer? Also find the maximum power. [Figure: a 12 V battery (positive terminal at top) on the left, connected across a 4 Ω resistor (top node P to bottom line); a 2 Ω resistor from node P to node Q; from node Q to the bottom line a branch of 6 Ω in series with a 10 V battery (positive terminal toward the 6 Ω, i.e. upward); load RL = 2 kΩ connected from node Q to the bottom line.] ## Short answer Using Thevenin’s theorem, Vth = 11.5 V, Rth = 1.5 Ω. With RL = 2 kΩ, the load current is IL = Vth/(Rth+RL) ≈ 5.75 mA. Maximum power transfer requires RL = Rth = 1.5 Ω, and the maximum power is Pmax = Vth²/(4Rth) = 22.04 W. ## Answer **Thevenin's theorem:** any linear two-terminal network can be replaced by a voltage source Vth in series with a resistance Rth, where Vth is the open-circuit voltage at the terminals and Rth is the resistance seen from the terminals with independent sources replaced by their internal resistances. Take the bottom line as reference. Node P is fixed at 12 V by the battery (ideal), so the 4 Ω across the battery does not affect the rest of the circuit. ```text P 2Ω Q o--/\/\---o----+-----+ | | | | 12V 4Ω 6Ω RL (open for Vth) | | | | +10V | o---------o----+ ``` #### Step 1: Thevenin voltage (RL removed) Current flows from P (12 V) through 2 Ω and 6 Ω into the + terminal of the 10 V battery: ```text I = (12 − 10)/(2 + 6) = 2/8 = 0.25 A Vth = VQ = 12 − I × 2 = 12 − 0.5 = 11.5 V check: VQ = 10 + 0.25 × 6 = 11.5 V ``` #### Step 2: Thevenin resistance Short both batteries. The 4 Ω is shorted by the 12 V source, so from Q we see 2 Ω in parallel with 6 Ω: ```text Rth = (2 × 6)/(2 + 6) = 1.5 Ω ``` #### Step 3: Load current (RL = 2 kΩ) ```text IL = Vth/(Rth + RL) = 11.5/(1.5 + 2000) = 5.746 × 10⁻³ A ``` **Answer: IL ≈ 5.75 mA** (power in the 2 kΩ load ≈ 66 mW). #### Step 4: Maximum power transfer Maximum power is delivered when RL = Rth: ```text RL = Rth = 1.5 Ω Pmax = Vth²/(4 Rth) = (11.5)²/(4 × 1.5) = 132.25/6 = 22.04 W ``` **Answer: RL = 1.5 Ω for maximum power; Pmax = 22.04 W.** The equivalent circuit is 11.5 V in series with 1.5 Ω feeding RL. With RL = 2 kΩ the load is badly mismatched, so it takes only about 0.3% of the maximum possible power. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does Thevenin's theorem say about a linear two-terminal network? A. It can be replaced by a current source in parallel with a resistance only B. It can be replaced by a voltage source Vth in series with Rth C. It can be replaced by a single resistor equal to the open-circuit voltage D. It can be replaced by an ideal short circuit Answer: B. Vth is the open-circuit voltage and Rth the resistance seen with sources deactivated. (Key point: Thevenin theorem statement) 2. Why does the 4 ohm resistor across the 12 V battery not affect the Thevenin equivalent? A. The 4 ohm is in series with RL B. The 4 ohm is open circuited by RL C. The ideal battery fixes node P at 12 V, so the 4 ohm only draws its own current D. It cancels the 10 V battery Answer: C. An ideal source holds its voltage regardless of parallel elements, so the 4 ohm does not change the rest of the circuit. (Key point: Effect of the 4 ohm resistor) 3. What are Vth and Rth at node Q? A. Vth = 12 V, Rth = 8 ohm B. Vth = 11.5 V, Rth = 8 ohm C. Vth = 10 V, Rth = 1.5 ohm D. Vth = 11.5 V, Rth = 1.5 ohm Answer: D. I = (12-10)/(2+6) = 0.25 A, Vth = 12 - 0.5 = 11.5 V; Rth = 2 || 6 = 1.5 ohm with batteries shorted. (Key point: Thevenin voltage and resistance) 4. For maximum power transfer in this circuit, what should RL be and why does RL = 2 kohm deliver little? A. RL = Rth = 1.5 ohm; 2 kohm is badly mismatched B. RL = Vth = 11.5 ohm; 2 kohm is too small C. RL = 2 kohm is the best match D. RL must be zero for maximum power Answer: A. Maximum power is delivered when RL = Rth (1.5 ohm), giving 22.04 W; 2 kohm gets about 0.3% of that. 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