The current divider is a shortcut for finding branch current in a parallel network. It converts a two-step problem into a single-step equation, saving us the effort of first solving for the source voltage.
Three common forms of the current divider equation are presented along with their derivation. Worked examples are included.
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Last updated: 25 November, 2025
Figure 1: Constant current source and parallel resistors used to demonstrate the current divider. 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.
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 two parallel resistor equation. It’s mathematically correct, but challenging as we must clearly differentiate between two closely related concepts:
- the equivalent parallel resistance of the circuit excluding the resistor of interest
- the equivalent resistance of the entire parallel circuit
Students often use the resistance of the entire circuit when they should have excluded the resistor of interest.
Tech Tip: If you do use this form, make two circles. One for the resistor of interest and one for the rest of the circuit.
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
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.
Define R1 as the 5 Ω resistor and R2 as the 7 Ω 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_1}{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 for 2A when the supply current is 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 circuits 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.”
Be sure to test your understanding using the associated learning companion.
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Best wishes,
APDahlen
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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 (interwoven). 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, completing a decades-long journey that began as a search for capacitors. Read his story here.