---
title: "Series-parallel resistors, find fourth resistance"
subject: "Basic Electrical Engineering"
chapter: 1
key: "semester_1/basic_electrical_engineering:series-parallel-resistors-fourth-resistance"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/series-parallel-resistors-fourth-resistance
site: BCT Engineering
---

# Series-parallel resistors, find fourth resistance

Basic Electrical Engineering · Chapter 1: General Electric System

- Asked in: 2074 Asoj, 5 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/series-parallel-resistors-fourth-resistance
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-1

## Question

A circuit containing three resistors with resistances 12Ω, 18Ω and 36Ω respectively joined in parallel is connected in series with a fourth resistance. The whole circuit is supplied at 60V and it is found that power dissipated in 12Ω resistance is 36watt. Determine the value of fourth resistance and the total power dissipated in the group.

## Answer

The 12 Ω, 18 Ω and 36 Ω resistors are in parallel, and this group is in series with the unknown resistor R₄ across 60 V.

#### Step 1: Voltage across the parallel group
P₁₂ = V²/12 = 36, so V² = 432 and V = 20.78 V

#### Step 2: Branch currents
- I₁₂ = 20.78/12 = 1.732 A
- I₁₈ = 20.78/18 = 1.155 A
- I₃₆ = 20.78/36 = 0.577 A

Total current I = 1.732 + 1.155 + 0.577 = 3.464 A

#### Step 3: Fourth resistance
Voltage across R₄ = 60 − 20.78 = 39.22 V

R₄ = 39.22/3.464 = 11.32 Ω

#### Step 4: Power in the group
- P₁₂ = 36 W
- P₁₈ = 20.78²/18 = 24 W
- P₃₆ = 20.78²/36 = 12 W

Power dissipated in the parallel group = 36 + 24 + 12 = 72 W

(Check: the whole circuit takes 60 × 3.464 = 207.8 W, of which R₄ takes 39.22 × 3.464 = 135.8 W.)

**Answer: R₄ ≈ 11.32 Ω; power in the group = 72 W**

## Check yourself

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

1. How is the voltage across the parallel group found from the 36 W in the 12 ohm resistor?
   A. V = P x R = 432 V
   B. V = sqrt(P x R) = 20.78 V
   C. V = P/R = 3 V
   D. V = 60 V, the full supply
   Answer: B. P = V^2/R gives V^2 = 36 x 12 = 432, so V = 20.78 V. (Key point: Voltage across the group)

2. What is the total current through the fourth resistor?
   A. 1.732 A, the current in the 12 ohm only
   B. 3.464 A, the sum of all three branch currents
   C. 0.577 A, the smallest branch current
   D. 5.196 A
   Answer: B. The group is in series with R4, so R4 carries the sum 1.732 + 1.155 + 0.577 = 3.464 A. (Key point: Total current)

3. R4 equals:
   A. 20.78/3.464 = 6 ohm
   B. 60/3.464 = 17.3 ohm
   C. (60 - 20.78)/3.464 = 11.32 ohm
   D. (60 - 20.78)/1.732 = 22.6 ohm
   Answer: C. R4 drops 60 - 20.78 = 39.22 V while carrying 3.464 A, so R4 = 11.32 ohm. (Key point: Fourth resistance)

4. What is the total power in the parallel group?
   A. 36 W
   B. 60 W
   C. 135.8 W
   D. 72 W
   Answer: D. Powers are 36 + 24 + 12 = 72 W (equivalently 20.78 x 3.464). (Key point: Power in the group)

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