---
title: "Find RL for maximum power (delta network)"
subject: "Basic Electrical Engineering"
chapter: 3
key: "semester_1/basic_electrical_engineering:max-power-rl-10v-20-60-90"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/max-power-rl-10v-20-60-90
site: BCT Engineering
---

# Find RL for maximum power (delta network)

Basic Electrical Engineering · Chapter 3: Network Theorems

- Asked in: 2070 Chaitra, 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-20-60-90
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3

## Question

For the circuit shown in figure below, what will be the value of RL to get the maximum power? What is the maximum power delivered to the load? [Figure: 10 V source; 20 Ω, 60 Ω and 90 Ω resistors arranged in a delta (triangle) network; load RL on the right.]

## Short answer

Find Thevenin equivalent: Rth = 20 ∥ 90 ∥ 60 = 12.86 Ω, Vth = 10 × Rth/20 = 6.43 V. For maximum power, RL = Rth = 12.86 Ω, and Pmax = Vth²/(4Rth) = 0.804 W.

## Answer

#### Reading the figure
All four elements 20 Ω (in series with the 10 V battery), 90 Ω, 60 Ω and R_L connect between the apex node T and the bottom rail. So, as seen by R_L, the rest of the circuit is a 10 V source with 20 Ω in series, in parallel with 90 Ω and 60 Ω (the drawing as a triangle is only the layout).

```text
 T o------+--------+--------+-------+
          |        |        |       |
        [20 Ω]    [90 Ω]  [60 Ω]  [R_L]
         (10 V)    |        |       |
 o-------+---------+--------+-------+
```

#### Step 1: Thevenin resistance (source shorted)
R_th = 20 ∥ 90 ∥ 60

1/R_th = 1/20 + 1/90 + 1/60 = 0.05 + 0.01111 + 0.01667 = 0.07778

R_th = **12.86 Ω** (90/7 Ω)

#### Step 2: Thevenin voltage (R_L removed)
V_th = 10 × R_th/20 = 10 × 12.857/20 = **6.43 V**

(The 10 V source feeds R_th through 20 Ω as a divider; the 90 Ω and 60 Ω have no source.)

#### Step 3: Maximum power
R_L = R_th = **12.86 Ω**

P_max = V_th²/(4 R_th) = 6.4286²/(4 × 12.857) = 41.33/51.43 = **0.804 W**

**Answer: R_L = 12.86 Ω; P_max = 0.804 W**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the condition for maximum power transfer to the load?
   A. R_L = 2 R_th
   B. R_L = R_th
   C. R_L is as large as possible
   D. R_L = R_th/2
   Answer: B. Maximum power is delivered when the load resistance equals the Thevenin resistance. (Key point: Max power condition)

2. How is R_th found for this circuit?
   A. Short the 10 V source and combine 20, 90 and 60 ohm in parallel = 12.86 ohm
   B. Take 20 ohm alone
   C. Open the 10 V source and take 90 ohm in series with 60 ohm
   D. Add 20 + 90 + 60 = 170 ohm
   Answer: A. With the source shorted, all three resistors lie between the same two nodes, so 1/R_th = 1/20 + 1/90 + 1/60, giving 12.86 ohm. (Key point: Finding R_th)

3. Why is V_th = 10 x R_th/20 = 6.43 V?
   A. The 10 V appears fully across R_L
   B. The 10 V is divided between the 20 ohm and R_th (the parallel 90 and 60 ohm)
   C. The 90 ohm and 60 ohm contain sources
   D. Because the load is shorted
   Answer: B. With R_L removed, the 10 V source feeds the 20 ohm in series with the parallel combination of 90 and 60 ohm; 90 || 60 = 36 ohm and 10 x 36/56 = 6.43 V, the same result. (Key point: Finding V_th)

4. Which expression gives P_max and what is its value here?
   A. V_th^2/R_th = 3.2 W
   B. V_th^2/(4 R_th) = 0.804 W
   C. V_th^2/(2 R_th) = 1.6 W
   D. V_th R_th = 82 W
   Answer: B. P_max = V_th^2/(4 R_th) = 6.43^2/(4 x 12.86) = 0.804 W. (Key point: Maximum power formula)

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