---
title: "Nodal analysis currents in all resistors"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:nodal-currents-23a-2a-10-ohm"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/nodal-currents-23a-2a-10-ohm
site: BCT Engineering
---

# Nodal analysis currents in all resistors

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2068 Shrawan, 8 marks
- Video: [Nodal Analysis for Circuits Explained](https://www.youtube.com/watch?v=f-sbANgw4fo) — 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/nodal-currents-23a-2a-10-ohm
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

Using Nodal Analysis method to find the current in individual resistors of the circuit shown below. [Figure: 25 A current source on the left; 2 Ω shunt; 2 Ω and 1 Ω in series along the middle of the top with a 10 Ω resistor bypass above; 5 Ω shunt; 4 Ω shunt; 2 A current source on the right pointing down.]

## Short answer

Node voltages: V_A = 32 V, V_B = 17.6 V, V_C = 13.93 V. Resistor currents: 16.0 A (2 Ω to ground), 7.20 A (A–B), 1.81 A (10 Ω A–C), 3.52 A (5 Ω), 3.68 A (1 Ω B–C), 3.48 A (4 Ω).

## Answer

#### Reading the figure
The source on the left is **25 A**, as stated in the question. The bottom rail is the reference (0 V).
- Node A: 25 A source (into A), 2 Ω to ground, 2 Ω to node B, 10 Ω (top bypass) to node C.
- Node B: 5 Ω to ground, 1 Ω to node C.
- Node C: 4 Ω to ground, 2 A source taking current out of C (arrow down).

```text
        +--------[10 Ω]--------------+
        |                            |
 A ---[2 Ω]--- B ---[1 Ω]--- C       |
 |  |           |             |      |
25A [2]        [5]           [4]   (2A)
 |  |           |             |      |
 +--+-----------+-------------+------+
```

#### Nodal equations (currents leaving each node)
Node A: V_A/2 + (V_A − V_B)/2 + (V_A − V_C)/10 = 25
Node B: (V_B − V_A)/2 + V_B/5 + (V_B − V_C)/1 = 0
Node C: (V_C − V_A)/10 + (V_C − V_B)/1 + V_C/4 + 2 = 0

Multiplying through (by 10, 10 and 20):
- 11 V_A − 5 V_B − V_C = 250
- −5 V_A + 17 V_B − 10 V_C = 0
- −2 V_A − 20 V_B + 27 V_C = −40

Solving (by Cramer's rule or elimination):
**V_A = 32.00 V, V_B = 17.60 V, V_C = 13.93 V**

#### Current in each resistor
| Resistor | Voltage | Current |
|---|---|---|
| 2 Ω (A to ground) | 32.00 V | 16.00 A |
| 2 Ω (A to B) | 14.39 V | 7.20 A (A to B) |
| 10 Ω (A to C) | 18.07 V | 1.81 A (A to C) |
| 5 Ω (B to ground) | 17.60 V | 3.52 A |
| 1 Ω (B to C) | 3.68 V | 3.68 A (B to C) |
| 4 Ω (C to ground) | 13.93 V | 3.48 A |

KCL check: at A, 16.00 + 7.20 + 1.81 = 25.0 A. At B, 7.20 = 3.52 + 3.68. At C, 1.81 + 3.68 = 3.48 + 2 (= 5.49 A).

**Answer: 16.0 A, 7.20 A, 1.81 A, 3.52 A, 3.68 A and 3.48 A as tabulated.**

(If the scan's source value is actually 28 A, the same equations give V_A = 36 V, V_B = 20 V, V_C = 16 V and currents 18, 8, 2, 4, 4 and 4 A.)

## Check yourself

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

1. How is the KCL equation written at each node in this answer?
   A. Sum of currents leaving the node through resistors equals the source current injected into it
   B. Sum of loop voltages equals zero
   C. Sum of resistances at the node equals the source current
   D. Node voltages are added and set equal to the supply voltage
   Answer: A. Each nodal equation sums the currents leaving the node through resistors (V difference / R) and equates it to the injected source current. (Key point: Nodal equation form)

2. What is taken as the reference (0 V) node?
   A. Node A
   B. The bottom rail
   C. Node C
   D. The top of the 10 ohm resistor
   Answer: B. The bottom rail is the common ground, so the three remaining nodes A, B, C are the unknowns. (Key point: Reference node)

3. What node voltages come out of the solution?
   A. V_A = 28 V, V_B = 20 V, V_C = 16 V
   B. V_A = 36 V, V_B = 16 V, V_C = 20 V
   C. V_A = 36 V, V_B = 20 V, V_C = 16 V
   D. V_A = 18 V, V_B = 8 V, V_C = 4 V
   Answer: C. Solving the three nodal equations gives 36, 20 and 16 V. (Key point: Node voltages)

4. What is the current in the 10 ohm resistor from A to C and why?
   A. 3.6 A, because 36/10
   B. 1.6 A, because 16/10
   C. 4 A, because (36 - 16)/5
   D. 2 A, because (36 - 16)/10
   Answer: D. Current in a resistor = voltage difference across it / R; 20 V across 10 ohm gives 2 A. (Key point: Branch current from node voltages)

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