---
title: "Equivalent resistance and power in 5Ω"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:equivalent-resistance-power-5-ohm"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/equivalent-resistance-power-5-ohm
site: BCT Engineering
---

# Equivalent resistance and power in 5Ω

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2068 Chaitra, 6 marks
- Video: [Equivalent Resistance of a Complex Circuit with Series and Parallel Resistors](https://www.youtube.com/watch?v=xn_Xnc97V4w) — Engineer4Free
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/equivalent-resistance-power-5-ohm
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

Find the equivalent resistance in the figure shown, and power dissipated in the 5Ω resistor. [Figure: supply voltage V; 1 Ω in series, then a parallel combination of: 2 Ω on top; 3 Ω in series with 4 Ω in the middle; 5 Ω at the bottom; followed by 6 Ω in series.]

## Short answer

Combine middle branch 3+4=7 Ω, parallel group 2,7,5 Ω gives 1.186 Ω, then add series 1 and 6 Ω: R_eq ≈ 8.19 Ω. Voltage across parallel group = 0.1449 V, power in 5 Ω: P = V²/5 = 0.0042 V² W (42 W for 100 V).

## Answer

The supply voltage *V* is not given on the scan, so the answer for the power is given in terms of *V*.

```text
        +--[2 Ω]------+
 o--[1]-+--[3]--[4]---+--[6]--o
        +--[5 Ω]------+
```

#### Equivalent resistance
Middle branch: 3 + 4 = 7 Ω

Parallel group: 1/R_p = 1/2 + 1/7 + 1/5 = 0.5 + 0.1429 + 0.2 = 0.8429
R_p = 1.186 Ω (70/59 Ω)

R_eq = 1 + 1.186 + 6 = **8.186 Ω**

#### Power in the 5 Ω resistor
Supply current: I = V/8.186 = 0.1222 V
Voltage across the parallel group: V_p = I × 1.186 = 0.1449 V
P₅ = V_p²/5 = (0.1449 V)²/5 = **0.0042 V² watts**

Example: for V = 100 V, I = 12.2 A, V_p = 14.49 V and P₅ = 42.0 W.

**Answer: R_eq ≈ 8.19 Ω; P₅ ≈ 0.0042 V² W (42 W for a 100 V supply)**

## Check yourself

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

1. What is the resistance of the middle branch?
   A. 7 ohm, since 3 and 4 are in series
   B. 12/7 ohm, since 3 and 4 are in parallel
   C. 1 ohm, the difference
   D. 3.5 ohm, the average
   Answer: A. 3 ohm in series with 4 ohm adds to 7 ohm. (Key point: Series branch)

2. How is the parallel group of 2, 7 and 5 ohm found?
   A. R_p = 2 + 7 + 5 = 14 ohm
   B. 1/R_p = 1/2 + 1/7 + 1/5, giving about 1.186 ohm
   C. R_p = 2 x 7 x 5 / (2+7+5)
   D. R_p = 1/2 + 1/7 + 1/5 = 0.843 ohm
   Answer: B. Parallel conductances add; R_p is the reciprocal of 0.8429, i.e. 1.186 ohm. Forgetting the final reciprocal is a common mistake. (Key point: Parallel group)

3. What is R_eq?
   A. 1.186 ohm
   B. 8.19 ohm excluding the 1 ohm
   C. 1 + 1.186 + 6 = 8.19 ohm
   D. 15 ohm
   Answer: C. The 1 ohm and 6 ohm are in series with the parallel group. (Key point: Total resistance)

4. How is the power in the 5 ohm resistor obtained?
   A. P = I^2 x 5 using the total supply current
   B. P = V^2/5 using the full supply voltage
   C. P = V^2/R_eq
   D. Find supply current, then voltage across the parallel group, then P = V_p^2/5
   Answer: D. The 5 ohm sees only the voltage across the parallel group, not the supply voltage or total current. (Key point: Power in the 5 ohm)

## Related in Basic Electrical Engineering

- [Series-parallel resistors, find fourth resistance](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-parallel-resistors-fourth-resistance) — Basic Electrical Engineering, chapter 1
- [Series-parallel circuit currents and power](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-parallel-z1-6-0-01h-100v) — Basic Electrical Engineering, chapter 6
- [Equivalent resistance between A and B](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aequivalent-resistance-24-8-8) — Basic Electrical Engineering, chapter 2
- [Total power of series-parallel circuit](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-parallel-total-power-265uf-19mh) — Basic Electrical Engineering, chapter 6
- [Unknown resistor from voltage across 500Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aunknown-resistor-r-500-ohm-2-5v) — Basic Electrical Engineering, chapter 1, asked 2×
- [Series path with parallel capacitor](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-rl-parallel-capacitor-240v) — Basic Electrical Engineering, chapter 6

## Related in other subjects

- [Mesh analysis: voltage across 1 Ω resistor](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Amesh-analysis-dependent-source-1ohm) — Electrical Circuit Theory, chapter 1
- [Series-parallel AC circuit currents and power](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Aseries-parallel-ac-circuit-power) — Fundamental of Electrical and Electronics Engineering, chapter 3
- [Power factor and series-parallel circuit calculation](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Apower-factor-series-parallel-circuit) — Fundamental of Electrical and Electronics Engineering, chapter 3
- [3 m wall under wind: fixed top and bottom](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Awind-wall-3m-fixed-top-bottom) — Concrete Technology and Masonry Structure, chapter 9, asked 2×