--- title: "Find RL for maximum power, 10 V network" subject: "Basic Electrical Engineering" chapter: 3 key: "semester_1/basic_electrical_engineering:max-power-rl-10v-2-ohm-network" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/max-power-rl-10v-2-ohm-network site: BCT Engineering --- # Find RL for maximum power, 10 V network Basic Electrical Engineering · Chapter 3: Network Theorems - Asked in: 2068 Shrawan, 8 marks - Video: [Maximum Power Transfer Theorem | Electric Circuits | Example 4.13 | Electrical Engineering](https://www.youtube.com/watch?v=x6b_DU7_8ro) — Electrical and Electronics Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/max-power-rl-10v-2-ohm-network - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3 ## Question Find the value of RL in the network shown below to obtain maximum power in it. Also find the maximum power in RL. [Figure: RL in the top branch; 2 Ω, 2 Ω in the top row; 2 Ω shunt in the middle; 10 V source on the left; 2 Ω in series with a 10 V source on the right.] ## Short answer Find Thevenin equivalent: Vth = 2 V, Rth = 1.2 Ω. For maximum power, RL = Rth = 1.2 Ω. Maximum power is Vth²/(4Rth) = 2²/(4×1.2) = 0.833 W. ## Answer #### Reading the figure Take the bottom rail as reference. R_L is connected between the top-left node L and the top-right node R. - Left: 10 V battery (+ on top) from the bottom rail to L. - 2 Ω from L to the middle node M; 2 Ω shunt from M to ground; 2 Ω from M to R. - Right: 2 Ω in series with a 10 V battery (+ on top, towards the 2 Ω) from R to ground. #### Step 1: Thevenin voltage across L-R (R_L removed) V_L = 10 V. With R_L removed, the only paths at R are the 2 Ω to M and the right branch (2 Ω with the 10 V battery): Node M: (V_M − 10)/2 + V_M/2 + (V_M − V_R)/2 = 0 → 3 V_M − V_R = 10 Node R: (V_R − V_M)/2 + (V_R − 10)/2 = 0 → −V_M + 2 V_R = 10 Solving: V_M = 6 V, V_R = 8 V V_th = V_L − V_R = 10 − 8 = **2 V** #### Step 2: Thevenin resistance (both batteries shorted) Shorting the left battery joins L to ground, so R_th is the resistance from R to ground: - Right branch: 2 Ω. - Path through M: 2 Ω (R to M) in series with M to ground, where the 2 Ω shunt is in parallel with the 2 Ω from M to L (2 ∥ 2 = 1 Ω): 2 + 1 = 3 Ω. - R_th = 2 ∥ 3 = **1.2 Ω** #### Step 3: Maximum power R_L = R_th = **1.2 Ω** P_max = V_th²/(4 R_th) = 2²/(4 × 1.2) = 4/4.8 = **0.833 W** **Answer: R_L = 1.2 Ω; P_max = 0.833 W** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For maximum power in R_L, R_L must equal what? A. The sum of all resistances in the circuit B. The Norton current C. The Thevenin resistance of the rest of the network seen from L-R D. Twice the Thevenin voltage Answer: C. Maximum power transfer occurs at R_L = R_th. (Key point: Condition and meaning) 2. What is V_th across L-R with R_L removed? A. V_L - V_R = 10 - 8 = 2 V B. 6 V C. 10 V D. 8 V Answer: A. Nodal equations give V_M = 6 V, V_R = 8 V; with V_L = 10 V, V_th = 2 V. (Key point: V_th) 3. With both batteries shorted, how is R_th found? A. R_th = 2 || 2 = 1 ohm B. R_th = 2 + 2 + 2 = 6 ohm C. R_th = 2 ohm (right branch) in parallel with 3 ohm = 1.2 ohm D. R_th = 3 ohm Answer: C. From R to ground the right 2 ohm is parallel to the path 2 + (2 || 2) = 3 ohm; 2 || 3 = 1.2 ohm. (Key point: R_th) 4. What is P_max? A. V_th^2/R_th = 3.33 W B. V_th^2/(2R_th) = 1.67 W C. V_th^2/(4R_th) = 4/4.8 = 0.833 W D. V_th x R_th = 2.4 W Answer: C. P_max = 2^2/(4 x 1.2) = 0.833 W. (Key point: P_max) ## Related in Basic Electrical Engineering - [Find RL for maximum power (delta network)](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Amax-power-rl-10v-20-60-90) — Basic Electrical Engineering, chapter 3 - [Find RL for maximum power and Pmax](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Amax-power-rl-8v-4v-6-ohm) — Basic Electrical Engineering, chapter 3 - [Find R for maximum power, 10 A and 48 V](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Amax-power-r-10a-48v) — Basic Electrical Engineering, chapter 3 - [Maximum power in bridge load RL](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Amax-power-bridge-rl-100v) — Basic Electrical Engineering, chapter 3 - [Maximum power transfer, find R](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Amax-power-transfer-6v-8v-2a) — Basic Electrical Engineering, chapter 3 - [Find R for maximum power to load](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Amax-power-load-4-ohm-find-r) — Basic Electrical Engineering, chapter 3 ## Related in other subjects - [Thevenin current through RL and maximum power](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Athevenin-max-power-load) — Fundamental of Electrical and Electronics Engineering, chapter 1 - [Superposition theorem and maximum power to RL](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Asuperposition-max-power-transfer) — Fundamental of Electrical and Electronics Engineering, chapter 1 - [Parallel RL and RC branch currents, power](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Aparallel-impedance-total-current) — Basic Electrical and Electronics Engineering, chapter 1 - [Driving point impedance and voltage ratio, bridged RL network](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Adriving-point-impedance-voltage-ratio-bridged) — Electrical Circuit Theory, chapter 5