What is the current divider?
The current divider is a mathematical shortcut used to solve for the branch current in a parallel circuit such as the one shown in Figure 1. We call it a shortcut as it converts a two-step problem into a single-step equation, saving us the effort of first solving for the source voltage.
This article presents the three most common forms of the current divider. It provides a brief introduction to the technique based on Ohm’s Law and Kirchhoff’s Law. We conclude with worked examples and a set of questions and critical thinking questions.
Figure 1: The current divider equation may be used to solve for the current flowing through any resistor as shown in this in this circuit. Ref: Multisim Live
Tech Tip: Click here for DigiKey’s easy to use online calculator to solve for the branch using the current divider equation.
What is the current divider equation?
There are three basic equations for the resistive current divider. The first is a generalized equation applicable to a parallel circuit with three or more resistors. The second is a straightforward equation used when only two parallel resistors are involved. The third equation, which we call the “error-prone equation”, is also applicable to a multi-resistor circuit. It is included to round out the article. However, IMO, this equation should not be added to your study notecards.
Generalized equation
I_{R_X} = I_{Total} \left( \frac{R_{Total}}{R_X} \right)
Where:
- I_{R_X} is the current through the resistor of interest
- I_{Total} is the total current
- R_{Total} is the equivalent resistance of the parallel resistor combination
- R_X is the resistor of interest
Two parallel resistors equation
I_{R_1} = I_{Total} \left( \frac{R_2}{R_1 + R_2} \right)
I_{R_2} = I_{Total} \left( \frac{R_1}{R_1 + R_2} \right)
Where:
- I_{R_1} is the current through resistor R_1
- I_{R_2} is the current through resistor R_2
- I_{Total} is the total current
- R_1 and R_2 are the resistances of the respective branches
Error-prone equation
Occasionally we encounter a third variation that attempts to generalize the previously mentioned two parallel resistor equation. Personally, I do not like this equation as it requires a picture to explain. It also adds an unnecessary cognitive load, requiring us to differentiate between the “equivalent parallel circuit resistance” and the “equivalent parallel resistance of the circuit excluding the resistor of interest.
The variation is included here for the sake of completing the article.
I_{R_X} = I_{Total} \left(\frac{R_{Total}}{R_{X}+R_{Total}}\right)
Where:
- I_{R_X} is the current through the resistor of interest
- I_{Total} is the total current
- R_{Total} is the equivalent resistance of the parallel combination excluding R_X
- R_X is the resistor of interest
Tech Tip: Recommend you avoid the error-prone equation. I’ve seen too many students confuse the related but subtle distinction between “equivalent parallel circuit resistance” and “equivalent parallel resistance of the circuit excluding the resistor of interest.” We will not address this error prone formula again.
Derivation of the general current divider equation
The generalized equation is easy to solve. It starts with an assertion of Kirchhoff’s Voltage Law (KVL) which tells us that all elements in a parallel circuit have the same node voltage.
I_{Total} = \frac{V_{Supply}} {R_{Total}}
I_{R_X} = \frac{V_{Supply}} {R_X}
Since the supply voltage is the same for all resistors in a parallel circuit:
I_{Total} \cdot R_{Total} = I_{R_X} \cdot R_X
This can be rearranged into our generalized equation:
I_{R_X} = I_{Total} \left( \frac{R_{Total}}{R_X} \right)
Derivation of the two-resistor equation
We could derive the two-resistor equation from the general equation. For example, here is the starting point for solving for I_{R_1}:
I_{R_1} = I_{Total} \left( \frac{R_{Total}}{R_X} \right) = I_{Total} \left( \frac{\left( \frac{1}{R_{1}} + \frac{1}{R_{2}} \right)^{-1}}{R_1} \right)
With a bit of algebra, we can simplify the equation into the form:
I_{R_1} = I_{Total} \left( \frac{R_2}{R_1 + R_2} \right)
The algebraic steps are part of your homework assignment in questions located at the end of this note.
Worked examples
We will now present three worked examples.
Example 1
Given a parallel connected circuit consisting of a 10 A current source with 5 Ω and 7 Ω resistors, find the current through each resistor.
For the 5 Ω resistor: I_{R_1} = I_{Total} \left( \frac{R_2}{R_1 + R_2} \right) = 10 \left( \frac{7}{5 + 7} \right) \approx 5.83 A
For the 7 Ω resistor: I_{R_2} = I_{Total} \left( \frac{R_2}{R_1 + R_2} \right) = 10 \left( \frac{5}{5 + 7} \right) \approx 4.17 A
Example 2
Solve for I_{R_2} from figure 1:
R_{Total} = \left( \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right)^{-1} = \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right)^{-1} = \frac{12}{13} \Omega
I_{R_2} = I_{Total} \left( \frac{R_{Total}}{R_X} \right) = 13 \left( \frac{\frac{12}{13}}{3} \right) = 4 A
Tech Tip: Don’t forget the inverse operation when solving for the total resistance.
Example 3
Given Figure 1, select a new R3 so that its current is 2 A when the supply current is 18 A.
2 = 18 \left( \frac{\frac{1}{\frac{1}{2} + \frac{1}{3} + \frac{1}{R_3}}}{R_3} \right)
After simplification, we determine that the appropriate resistor is 9.6 Ω as seen in Figure 2. The algebra is left as a homework assignment.
Figure 2: Select a value for R3 that will increase the current to 18 A.
Conclusion
The current divider is a shortcut designed to save time while calculating the branch current in a parallel circuit. There are many ways to describe the current divider. The typical textbook presents a two-resistor solution and then includes a generalized formula for multiple resistor calculations. If you must select one and only one formula, I recommend you use:
I_{R_X} = I_{Total} \left( \frac{R_{Total}}{R_X} \right)
It’s a straightforward formula that will accommodate small circuit with two resistors and complex circuits with hundreds of parallel branches. While the other formulas work, the two-resistor formula is too specialized and the other is too complex requiring additional mental capacity to differentiate “equivalent parallel circuit resistance” from the “equivalent parallel resistance of the circuit excluding the resistor of interest.”
Best wishes,
APDahlen
Helpful links
Please follow these links to related and useful information:
- Digikey’s product selection guides
- Voltage Divider Fundamentals
- Voltage Divider Calculations Involving Three or More Resistors
- Voltage Divider Conversion Calculator
- How to select resistor pairs for op amp and voltage divider applications
About this author
Aaron Dahlen, LCDR USCG (Ret.), serves as an application engineer at DigiKey. He has a unique electronics and automation foundation built over a 27-year military career as a technician and engineer which was further enhanced by 12 years of teaching (partially interwoven with military experience). With an MSEE degree from Minnesota State University, Mankato, Dahlen has taught in an ABET-accredited EE program, served as the program coordinator for an EET program, and taught component-level repair to military electronics technicians. Dahlen has returned to his Northern Minnesota home and thoroughly enjoys researching and writing educational articles about electronics and automation.
Highlighted Experience
Dahlen is an active contributor to the DigiKey TechForum. At the time of this writing, he has created over 170 unique posts and provided an additional 500 forum posts. Dahlen shares his insights on a wide variety of topics including microcontrollers, FPGA programming in Verilog, and a large body of work on industrial controls.
Questions
The following questions will help reinforce the content of the article.
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Present the two recommended forms of the current divider equations.
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Present and then solve example problems for each equation. Be sure to include example involving multiple resistors. Be sure to work the problem backwards looking for th resistance given a desired supply and branch current.
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What are the properties of an ideal constant current source?
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What are the properties of an ideal constant voltage source?
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Assuming ideal components for Figure 1, what is the voltage developed across a current source when the resistors are:
A) open circuited
B) short circuited -
Show the algebraic steps to convert the general current divider equation into a two-resistor solution for I_{R_2}.
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Show the algebraic steps required to solve the 3rd worked example.
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Given Figure 1, select a new R2 so that the R2 branch current is 5A when the supply current is 15 A. Support your choice by showing your work.
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Contrast and compare the current divider equations and the voltage divider equations. Hint: Construct a table showing the resistor pair equations as well as the multi resistor equations. Look for patterns that will help you remember all of the equations.
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Research and then present the picture for the error prone equation. Show how to use the equation to solve for the Figure 1 branch current for R3.
Critical thinking questions
These critical thinking questions expand the article’s content allowing you to develop a big picture understanding the material and its relationship to adjacent topics. They are often open ended, require research, and are best answered in essay form.
11) There is a tension in this article with regards to the “error prone equation.” Describe the reasoning for the author’s apprehension. Do you personally agree with the author? Where is the point of confusion and how can it be avoided?
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Example 3 presents an interesting property of current dividers. Explain why you could or could not solve the problem if the instructions had been shortened to “Given Figure 1, select a new R3 for a supply current of 18 A.” Hint: Does the current source depend on the value of R_3?
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What is a Thévenin equivalent circuit? How could it be used to simplify Figure 1 especially when solving for the R3 branch current.
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What passive electrical component has constant current properties? Hint: The comparison is transient and only applicable for the moment the device is turned off.