---
title: "In a direct current circuit, a main line carrying a constant current \\(I_0\\) reaches a junction that splits into three parallel branches, which rejoin at a second junction.  – Path 1 consists of a single resistor of resistance \\(R\\). – Path 2 consists of two resistors, each of resistance \\(R\\), connected in series. – Path 3 consists of two resistors, each of resistance \\(R\\), connected in parallel with each other.  Which of the following correctly ranks the currents \\(I_1\\), \\(I_2\\), and \\(I_3\\) through the three paths and provides a valid justification?"
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url: "https://nerd-notes.com/ubq/118358/"
date_modified: "2026-08-04T08:09:40+00:00"
---

# In a direct current circuit, a main line carrying a constant current \(I_0\) reaches a junction that splits into three parallel branches, which rejoin at a second junction.

– Path 1 consists of a single resistor of resistance \(R\).
– Path 2 consists of two resistors, each of resistance \(R\), connected in series.
– Path 3 consists of two resistors, each of resistance \(R\), connected in parallel with each other.

Which of the following correctly ranks the currents \(I_1\), \(I_2\), and \(I_3\) through the three paths and provides a valid justification?

In a direct current circuit, a main line carrying a constant current \(I_0\) reaches a junction that splits into three parallel branches, which rejoin at a second junction.

- Path 1 consists of a single resistor of resistance \(R\).
- Path 2 consists of two resistors, each of resistance \(R\), connected in series.
- Path 3 consists of two resistors, each of resistance \(R\), connected in parallel with each other.

Which of the following correctly ranks the currents \(I_1\), \(I_2\), and \(I_3\) through the three paths and provides a valid justification?

![A schematic diagram showing a horizontal circuit line carrying current I_0 from the left into a vertical junction line. The junction line branches into three horizontal parallel paths that extend to the right and rejoin at a second vertical line leading into a single horizontal output wire. Path 1 (top) contains a single resistor labeled R with an arrow showing current I_1. Path 2 (middle) contains two resistors in series, each labeled R, with an arrow showing current I_2. Path 3 (bottom) contains a parallel sub-loop of two resistors, each labeled R, with an arrow showing current I_3 entering Path 3. All current arrows point to the right. No other labels, text, or symbols appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1785830980-JCcyZk.jpg)

- **A.** \(I_2 > I_1 > I_3\), because Path 2 contains the greatest total number of resistors, which allows it to pull the largest share of current from the junction.
- **B.** \(I_1 = I_2 = I_3\), because Kirchhoff's junction rule dictates that the total current \(I_0\) entering a junction must divide equally among all available parallel paths.
- **C.** \(I_3 > I_1 > I_2\), because all three parallel paths experience the same potential difference, making the current in each path inversely proportional to its equivalent resistance.
- **D.** \(I_3 > I_1 > I_2\), because the potential drop across Path 3 is greater than the potential drop across Path 1, driving a greater current through Path 3.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/118358/*
