---
title: "Resistors A, B in parallel with C in series"
subject: "Basic Electrical Engineering"
chapter: 1
key: "semester_1/basic_electrical_engineering:resistors-a-b-c-4320w"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/resistors-a-b-c-4320w
site: BCT Engineering
---

# Resistors A, B in parallel with C in series

Basic Electrical Engineering · Chapter 1: General Electric System

- Asked in: 2073 Chaitra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/resistors-a-b-c-4320w
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-1

## Question

A direct current circuit comprises two resistors A of value 25Ω and B of unknown value, connected in parallel together with a third resistor C of value 5Ω connected in series with the parallel group. The potential difference across C is found to be 90V. If the total power in the circuit is 4320 watt. Calculate (i) the value of resistor B, (ii) the voltage applied to the ends of the whole circuit, (iii) the current in each resistor.

## Short answer

Total current I = 90V/5Ω = 18A. Applied voltage V = 4320W/18A = 240V. Voltage across parallel pair = 240V − 90V = 150V. I_A = 150V/25Ω = 6A, I_B = 18A − 6A = 12A. R_B = 150V/12A = 12.5Ω. (i) 12.5Ω, (ii) 240V, (iii) I_A=6A, I_B=12A, I_C=18A.

## Answer

A (25 Ω) and B are in parallel; this group is in series with C (5 Ω).

```text
 o--+--[A 25]--+--[C 5]--o
    +--[B ?]---+
```

#### Step 1: Total current
I = V_C/R_C = 90/5 = 18 A

#### Step 2: Applied voltage
P = V I, so V = 4320/18 = 240 V

#### Step 3: Voltage across the parallel group
V_p = 240 − 90 = 150 V

#### Step 4: Currents
- I_A = 150/25 = 6 A
- I_B = 18 − 6 = 12 A
- I_C = 18 A

#### Step 5: Resistor B
R_B = 150/12 = 12.5 Ω

(Check: 18² × 5 + 6² × 25 + 12² × 12.5 = 1620 + 900 + 1800 = 4320 W.)

**Answer: (i) R_B = 12.5 Ω; (ii) V = 240 V; (iii) I_A = 6 A, I_B = 12 A, I_C = 18 A**

## Check yourself

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

1. How is the total current found from the 90 V across C = 5 ohm?
   A. I = 4320/5 = 864 A
   B. I = 90 x 5 = 450 A
   C. I = 90/5 = 18 A, since C carries the total current
   D. I = 4320/90 = 48 A
   Answer: C. C is in series with the parallel group, so it carries the total current: 90/5 = 18 A. (Key point: Total current)

2. The applied voltage is found from:
   A. V = P x I = 77760 V
   B. V = P/I = 4320/18 = 240 V
   C. V = 90 + 25 = 115 V
   D. V = I x 5 = 90 V
   Answer: B. Total power P = V I, so V = 4320/18 = 240 V. (Key point: Applied voltage)

3. What is the voltage across the parallel group A and B, and the current in A?
   A. 240 V; 9.6 A
   B. 90 V; 3.6 A
   C. 150 V; 6 A
   D. 150 V; 12 A
   Answer: C. V_p = 240 - 90 = 150 V, so I_A = 150/25 = 6 A. (Key point: Parallel group)

4. Resistor B is:
   A. 12.5 ohm, since I_B = 18 - 6 = 12 A and V = 150 V
   B. 25 ohm
   C. 8.33 ohm using 150/18
   D. 15 ohm using 90/6
   Answer: A. I_B = 18 - 6 = 12 A, so R_B = 150/12 = 12.5 ohm. (Key point: Resistor B)

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