Current Division in Parallel Circuits

Recommended resources

The freely available resources this lesson will direct you to.

Hello! Let's continue our journey through circuit analysis.

In our last lesson, we mastered the Voltage Divider Rule, discovering how voltage in a series circuit splits proportionally to resistance. We built the intuition that a larger resistance requires a larger voltage "push" to maintain the same current.

Today, we'll explore the dual concept for parallel circuits. Just as voltage divides in series, current divides in parallel. This leads us to the Current Divider Rule (CDR). Your goal for this lesson is to apply the current divider rule to parallel circuits and explain intuitively why current splits inversely with resistance. This is another essential shortcut that strengthens your physical reasoning about circuit behavior.

1. The Intuition: Why Current Divides Inversely

Before diving into formulas, let's build the core intuition. Remember the two fundamental properties of a parallel circuit:

The voltage is the same across all parallel branches.
The total current entering the parallel combination must split and flow through the branches (Kirchhoff's Current Law).

If every parallel branch has the same voltage "pushing" the charge, where will the most charge flow? It will flow where it's easiest—through the path of least resistance. This is the fundamental reason why current and resistance have an inverse relationship in a parallel circuit.

A common and effective analogy is water flowing through pipes.

Current Division Rule: A Practical Walkthrough From an Engineer ...

The article 'Current Division Rule: A Practical Walkthrough From an Engineer' provides a great mental model for this concept. We'll start there to build your intuition.

Read the section titled 'A Quick Mental Picture Before the Math'. It uses a simple water pipe analogy that makes the inverse relationship between current and resistance feel obvious.

To solidify this idea, let's consider two extreme cases: a short circuit (zero resistance) and an open circuit (infinite resistance). Thinking about these edge cases is a powerful problem-solving technique in engineering and programming.

☑️10 - Parallel Resistors and Current Division

This video from SkanCity Academy explores how current behaves when one of the parallel paths is a short or open circuit. This vividly demonstrates the principle of current following the path of least resistance.

Watch the segment from 07:27 to 11:14. Observe how in the case of a short circuit (R=0), all the current bypasses the other resistor. Conversely, in an open circuit (R=∞), no current flows through that path, forcing all of it through the other resistor.

These examples confirm our intuition: current will always favor the path that is easier to traverse. The higher the resistance, the less current will flow through that path, given the same voltage across it.

2. Deriving the Current Divider Formula

Now that you have a solid intuitive grasp, let's translate this physical behavior into a precise mathematical formula. The current divider rule can be derived directly from Ohm's Law and Kirchhoff's Current Law, which is a great way to see how these fundamental principles connect.

Let's look at the derivation for two resistors in parallel.

Resistors in Parallel with Current Division — A total current, $i_{total}$, enters a node and splits into two branch currents, $i_1$ and $i_2$, through parallel resistors $R_1$ and $R_2$.

The following video walks through the algebra step-by-step.

☑️10 - Parallel Resistors and Current Division

The first part of the SkanCity Academy video provides a clear derivation of the current divider rule for a two-resistor circuit. This connects the foundational laws to the final shortcut formula.

Watch the video from the beginning to 07:27. Follow the derivation, which starts with the facts that voltage is equal across the branches (V = I_1R_1 = I_2R_2) and that the currents sum up (I_{total} = I_1 + I_2).

The derivation leads to the current divider formula for two parallel resistors:

I_{1} = I_{t o t a l} \cdot \frac{R _{2}}{R _{1} + R _{2}} and I_{2} = I_{t o t a l} \cdot \frac{R _{1}}{R _{1} + R _{2}}

Notice something interesting? To find the current in branch 1 ( $I_{1}$ ), we use the resistance from branch 2 ( $R_{2}$ ) in the numerator. This is the mathematical expression of the inverse relationship we've been discussing. The current in a given branch is proportional to the resistance of the other branch.

What about more than two resistors?

The two-resistor formula is a special case. The general formula for any number of parallel resistors is more cleanly expressed using conductance ( $G$ ), which is the reciprocal of resistance ( $G = 1/ R$ ). Conductance measures how easily current can flow.

With conductance, the relationship becomes proportional: current splits in direct proportion to the conductance of the branch.

I_{x} = I_{t o t a l} \cdot \frac{G _{x}}{G _{t o t a l}} = I_{t o t a l} \cdot \frac{1/ R _{x}}{\frac{1}{R _{1}} + \frac{1}{R _{2}} + \dots + \frac{1}{R _{N}}}

This video introduces and applies the general formula for a three-resistor circuit.

Current Dividers Explained!

The video 'Current Dividers Explained!' by The Organic Chemistry Tutor demonstrates the general formula for three or more resistors. This shows you how to handle more complex parallel networks.

Watch the segment from 03:55 to 06:38. Pay attention to how the formula is constructed using the reciprocals of the resistances (conductances).

3. Putting the Rule into Practice

Let's work through an example to solidify your understanding.

Current Divider Rule for Two Parallel Resistors — A circuit with a 36V source connected to two parallel resistors, $R_1 = 2 \text{ k}\Omega$ and $R_2 = 8 \text{ k}\Omega$. The image also provides the relevant formulas.

We want to find the currents $I_{1}$ and $I_{2}$ .

Step 1: Find the total current, $I_{T}$ .
First, we need the equivalent resistance $R_{e q}$ of the parallel combination.
$R_{e q} = \frac{R _{1} R _{2}}{R _{1} + R _{2}} = \frac{( 2 k Ω ) ( 8 k Ω )}{2 k Ω + 8 k Ω} = \frac{16 k Ω ^{2}}{10 k Ω} = 1.6 k Ω$
Now, use Ohm's Law to find the total current from the source.
$I_{T} = \frac{E}{R _{e q}} = \frac{36 V}{1.6 k Ω} = 22.5 mA$

Step 2: Apply the Current Divider Rule.
Now we use the total current $I_{T} = 22.5 mA$ and the CDR formulas.
To find $I_{1}$ (current through $R_{1}$ ), we use $R_{2}$ in the numerator:
$I_{1} = I_{T} \cdot \frac{R _{2}}{R _{1} + R _{2}} = 22.5 mA \cdot \frac{8 k Ω}{2 k Ω + 8 k Ω} = 22.5 mA \cdot \frac{8}{10} = 18 mA$
To find $I_{2}$ (current through $R_{2}$ ), we use $R_{1}$ in the numerator:
$I_{2} = I_{T} \cdot \frac{R _{1}}{R _{1} + R _{2}} = 22.5 mA \cdot \frac{2 k Ω}{2 k Ω + 8 k Ω} = 22.5 mA \cdot \frac{2}{10} = 4.5 mA$

Step 3: Verify the result.
A good habit is to check your work.

Check with KCL: Do the branch currents add up to the total current? $I_{1} + I_{2} = 18 mA + 4.5 mA = 22.5 mA$ . Yes, it matches $I_{T}$ .
Check with Intuition: Does more current flow through the smaller resistor? $R_{1}$ (2kΩ) is smaller than $R_{2}$ (8kΩ), and $I_{1}$ (18 mA) is indeed larger than $I_{2}$ (4.5 mA). The math confirms our physical intuition.

Your Turn: Practice Problem

Now it's your turn to apply the general formula.

A total current of 15A flows into a parallel combination of three resistors: $R_{1} = 2Ω$ , $R_{2} = 3Ω$ , and $R_{3} = 6Ω$ . Find the current flowing through each resistor ( $I_{1}$ , $I_{2}$ , and $I_{3}$ ).

Take a moment to solve it, then check your solution below.

Click here to see the solution

Solution:

We'll use the general formula based on conductance. First, let's find the total conductance, $G_{t o t a l}$ .
$G_{t o t a l} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}} = \frac{1}{2Ω} + \frac{1}{3Ω} + \frac{1}{6Ω}$
To add these, we find a common denominator (6):
$G_{t o t a l} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1 Siemens (S)$
Now we find the current in each branch:
$I_{1} = I_{t o t a l} \cdot \frac{1/ R _{1}}{G _{t o t a l}} = 15 A \cdot \frac{1/2 S}{1 S} = 7.5 A$
$I_{2} = I_{t o t a l} \cdot \frac{1/ R _{2}}{G _{t o t a l}} = 15 A \cdot \frac{1/3 S}{1 S} = 5 A$
$I_{3} = I_{t o t a l} \cdot \frac{1/ R _{3}}{G _{t o t a l}} = 15 A \cdot \frac{1/6 S}{1 S} = 2.5 A$
Check: $7.5 A + 5 A + 2.5 A = 15 A$ . The solution is consistent with KCL and our intuition (lowest resistance gets the highest current).

Conclusion

Today we've demystified how current divides in parallel circuits, adding the Current Divider Rule to your analytical toolbox.

Key Takeaways:

Intuition: Current divides among parallel paths by taking the path of least resistance. Because the voltage is the same across all branches, the current is inversely proportional to the resistance ( $I \propto 1/ R$ ).
Two-Resistor Formula: The current through one branch is the total current times the ratio of the other resistance to the total resistance: $I_{1} = I_{t o t a l} \cdot \frac{R _{2}}{R _{1} + R _{2}}$ .
General N-Resistor Formula: It's easiest to think in terms of conductance ( $G = 1/ R$ ). The current splits in direct proportion to the conductance of each branch: $I_{x} = I_{t o t a l} \cdot \frac{G _{x}}{G _{t o t a l}}$ .

Preview of the Next Lesson:
We have now mastered the two fundamental divider rules, which are powerful shortcuts. In the upcoming lessons, we will move from these specific patterns to a systematic and universal method for analyzing any circuit: Nodal Analysis. We will start by learning how to properly identify nodes and set up KCL equations, the very foundation of this powerful technique.