---
title: "Unknown resistor from voltage across 500Ω"
subject: "Basic Electrical Engineering"
chapter: 1
key: "semester_1/basic_electrical_engineering:unknown-resistor-r-500-ohm-2-5v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/unknown-resistor-r-500-ohm-2-5v
site: BCT Engineering
---

# Unknown resistor from voltage across 500Ω

Basic Electrical Engineering · Chapter 1: General Electric System

- Asked in: 2071 Shrawan, 6 marks; 2068 Shrawan, 10 marks (asked 2 times)
- Video: [Calculating an Unknown resistor explained.](https://www.youtube.com/watch?v=yZ-vVJuTDUk) — Digital School 101 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/unknown-resistor-r-500-ohm-2-5v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-1

## Question

What is the value of the unknown resistor 'R' in figure below, if the voltage drop across 500Ω resistor is 2.5 volts? [Figure: 550 Ω in series on top; then 50 Ω in the top branch; R and 500 Ω in parallel vertical branches.]

## Short answer

Assuming a 12 V supply, the voltage across the parallel group is 2.75 V. Current through the 500 Ω branch is 5 mA, total current is 16.82 mA, so the unknown R = 2.75 V / 11.82 mA ≈ 233 Ω.

## Answer

The scanned figure of one paper shows no supply value, but the same question in the 2068 Shrawan paper shows a **12 V** supply across the input terminals. I use 12 V as the assumed supply.

Circuit: 550 Ω in series with the parallel combination of R and (50 Ω + 500 Ω in series).

```text
 o--[550]--+--[50]--+
           |        |
          [R]     [500]
           |        |
 o---------+--------+
```

#### Step 1: Current in the 50 Ω and 500 Ω branch
I₅₀₀ = 2.5/500 = 5 mA

#### Step 2: Voltage across the parallel group (node voltage)
V_N = 2.5 + 5 mA × 50 = 2.5 + 0.25 = 2.75 V

#### Step 3: Current in the 550 Ω resistor
Voltage across 550 Ω = 12 − 2.75 = 9.25 V
I = 9.25/550 = 16.82 mA

#### Step 4: Current in R
I_R = 16.82 − 5 = 11.82 mA

#### Step 5: Value of R
R = 2.75/11.82 mA = 232.7 Ω

**Answer: R ≈ 233 Ω (for a 12 V supply)**

## Check yourself

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

1. How is the circuit arranged in this solution?
   A. 550 ohm in series with the parallel combination of R and (50 + 500) ohm in series
   B. 550 ohm in parallel with R, all in series with 50 and 500 ohm
   C. R, 50 ohm and 500 ohm all in parallel
   D. 550 ohm and R in series, parallel with 500 ohm
   Answer: A. R is in parallel with the 50 + 500 ohm series branch, and 550 ohm is in series with that group. (Key point: Circuit structure)

2. With 2.5 V across the 500 ohm resistor, what is the voltage across the parallel group (the node voltage)?
   A. 2.5 V
   B. 2.75 V
   C. 12 V
   D. 9.25 V
   Answer: B. I = 2.5/500 = 5 mA flows through 50 ohm too, dropping 0.25 V, so V_N = 2.5 + 0.25 = 2.75 V. (Key point: Node voltage)

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