---
title: "Source current by Kirchhoff laws"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:kirchhoff-source-current-polarity-bd-0-2a"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/kirchhoff-source-current-polarity-bd-0-2a
site: BCT Engineering
---

# Source current by Kirchhoff laws

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2069 Asar, 6 marks
- Video: [Electrical Engineering: Basic Laws (12 of 31) Kirchhoff's Laws: A Harder](https://www.youtube.com/watch?v=iY9OqWvB744) — Michel van Biezen
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/kirchhoff-source-current-polarity-bd-0-2a
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

Use Kirchhoff's laws to determine the magnitude of source current and polarity of the source S, if the current flowing through branch BD is 0.2A from D to B in the circuit shown below. [Figure: bridge with nodes A, B, C, D; AB = 25 Ω, BC = 50 Ω, BD = 100 Ω carrying 0.2 A from D to B, AD = 20 Ω, DC = 10 Ω; source S with a 5 Ω resistor in the path from D back to the supply.]

## Short answer

Source current is 3.52 A. Positive terminal of source S is on the C side, negative terminal towards the 5 Ω resistor and A.

## Answer

#### Reading the figure
Bridge: A–B = 25 Ω, B–C = 50 Ω, A–D = 20 Ω, D–C = 10 Ω, B–D = 100 Ω. The source S and a 5 Ω resistor are in series and connected between A and C (outside the bridge).

#### Step 1: Voltage across BD
Current 0.2 A flows from D to B, so V_D − V_B = 0.2 × 100 = 20 V.

Take V_A = 0 and let V_B = x, so V_D = x + 20 and V_C = c.

#### Step 2: KCL at B and D
At B (in from A, in from D, out to C):
(0 − x)/25 + 0.2 = (x − c)/50
⇒ c = 3x − 10 ... (1)

At D (in from A; out to B and to C):
(0 − (x + 20))/20 = 0.2 + (x + 20 − c)/10
⇒ c = 1.5 x + 32 ... (2)

From (1) and (2): 3x − 10 = 1.5 x + 32, so x = 28 V and c = 74 V.

So V_B = 28 V, V_D = 48 V, V_C = 74 V (with V_A = 0).

#### Step 3: Bridge currents
- B to A: 28/25 = 1.12 A
- D to A: 48/20 = 2.4 A
- C to B: (74 − 28)/50 = 0.92 A
- C to D: (74 − 48)/10 = 2.6 A

Current entering the bridge at C = 0.92 + 2.6 = 3.52 A, and the same leaves at A.

#### Step 4: Source current and polarity
The source branch must carry 3.52 A from A back to C, through the 5 Ω and S.

**Source current = 3.52 A**

Voltage needed: C is at +74 V relative to A. The 5 Ω drops 5 × 3.52 = 17.6 V, so the source must supply
E = 74 + 17.6 = 91.6 V

**Polarity:** the positive terminal of S is on the C side (towards C), the negative terminal is towards the 5 Ω and A. The source current is 3.52 A and the EMF is about 91.6 V.

## Check yourself

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

1. 0.2 A from D to B through 100 Ω gives
   A. V_B - V_D = 20 V
   B. V_D - V_B = 0.2 V
   C. V_D - V_B = 500 V
   D. V_D - V_B = 20 V
   Answer: D. V = IR = 0.2 x 100. (Key point: BD voltage)

2. Taking V_A = 0, the solution is
   A. V_B = 74 V, V_D = 48 V, V_C = 28 V
   B. V_B = 20 V, V_D = 0
   C. V_B = 28 V, V_D = 48 V, V_C = 74 V
   D. V_B = 28 V, V_D = 8 V, V_C = 74 V
   Answer: C. V_D = V_B + 20 = 48; from KCL c = 74. (Key point: Node voltages)

3. The positive terminal of S faces
   A. C, because C is at +74 V relative to A
   B. A
   C. B
   D. D
   Answer: A. C is the higher potential, so the source + terminal is on the C side. (Key point: Polarity)

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