Current Divider Formula and Derivation

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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 because 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.

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:stopwatch: Estimated reading time: 6 minutes

Figure 1: The current divider equation may be used to solve for the current flowing through any resistor in this circuit.
Figure 1: The current divider equation may be used to solve for the current flowing through any resistor in this circuit.975×537 61.6 KB

Figure 1: The current divider equation may be used to solve for the current flowing through any resistor 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, in my opinion, 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.

Define R1 as the 5 Ω resistor and R2 as her 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 that will increase the current to 18 A.
Figure 2: Select a value for R3 that will increase the current to 18 A.975×544 58.4 KB

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 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.