# Current Divider > A current divider splits a total current between two parallel resistors, and the smaller resistor always takes the bigger share. Page: https://formulabro.yunusemrevurgun.com/electricity/current-divider/ Category: Electricity Published: 2026-10-22 Updated: 2026-10-22 ## Formula $$ I_1 = I_{\text{total}} \frac{R_2}{R_1 + R_2} $$ Read it out loud: "I sub 1 equals I sub total times R sub 2, divided by the quantity R sub 1 plus R sub 2. Read I sub 1 as I sub one, the current through branch 1. I sub total is the total current that comes into the junction. R sub 1 is the first resistance and R sub 2 is the second. The word sub means the small label written low, so R sub 2 is R two." ## Before you start - **Current, the ampere (A) and the milliampere (mA)**: Current is the flow of electric charge past a point each second. Its unit is the **ampere**, written A. One ampere means one coulomb of charge passes a point every second. A **milliampere**, written mA, is one thousandth of an ampere, so 10 mA is 0.01 A. The prefix **milli** (m) always means divide by 1000. - **Resistance, the ohm ($\Omega$) and Ohm's law**: Resistance is how hard a part makes it for current to pass. Its unit is the **ohm**, written with the Greek capital letter omega, $\Omega$. Ohm's law says voltage equals current times resistance, $V = I R$, so for a fixed voltage the current is $I = V / R$. A bigger resistance lets less current through, and a smaller one lets more. - **Parallel connection**: Two parts are in **parallel** when both connect to the same two points, one wire on each side. Each part then sees the same voltage across it. Current that reaches the junction has a choice of paths, so it splits between them. This is different from a **series** connection, where one path carries all the current through every part. - **Subscripts, such as $R_1$ and $I_{\text{total}}$**: A small label written low, after a letter, is a **subscript**. It tells similar quantities apart without needing a new letter. $R_1$ is the resistance of branch 1 and $R_2$ is the resistance of branch 2. $I_{\text{total}}$ is the total current, with the word total written low. A subscript is only a name. It does not mean multiply. - **Fractions inside a formula**: A fraction bar means divide. The number above the bar, the **numerator**, is divided by the number below it, the **denominator**. In $\frac{R_2}{R_1 + R_2}$ the top is $R_2$ and the bottom is the total of both resistances. The answer is a plain number, and if the top is smaller than the bottom, the fraction is less than 1. More notation help: https://formulabro.yunusemrevurgun.com/math-basics/ ## What each symbol means - $I_1$: The **current through branch 1**, the branch we want to find. It is measured in amperes (A), where one ampere is one coulomb of charge per second. A small indicator light may draw only a few milliamperes, while a desk lamp draws a fraction of an ampere. - $I_{ ext{total}}$: The **total current** that enters the junction before it splits, measured in amperes (A). It is the sum of the currents in all the branches. In the example it is 10 mA, which is 0.01 A, a typical current for a small learning circuit. - $R_1$: The **resistance of branch 1**, measured in ohms ($\Omega$). One ohm is the resistance that lets one ampere flow when one volt is across it. Resistors used in small electronics usually range from a few ohms to millions of ohms. - $R_2$: The **resistance of branch 2**, the other path, measured in ohms ($\Omega$). It has the same unit and the same range as $R_1$. In the formula, $R_2$ sits on top of the fraction, so it controls how much current goes into branch 1. ## The formula piece by piece - $I_{\text{total}}$: Start with the current that arrives at the junction. This is the whole current that is about to be shared between the two branches. It is measured in amperes, and it sets the size of the amount being split, while the rest of the formula only decides how it is split. - $R_1 + R_2$: The two resistances added together. This sum is the whole that each share is measured against. Add the two resistances to get the total resistance of the pair, then use that total as the bottom of the share fraction. - $\frac{R_2}{R_1 + R_2}$: The **share** that goes into branch 1. It is a number between 0 and 1, with $R_2$ on top and the total on the bottom. If $R_2$ is much bigger than $R_1$, the fraction is close to 1, so almost all the current goes into branch 1. The other resistor sits on top because a bigger resistance in branch 2 pushes current away from branch 2. - $I_{\text{total}} \frac{R_2}{R_1 + R_2}$: Multiply the total current by the share. The total gives the amount and the fraction gives the portion. Multiplying a current by a fraction takes that fraction of it, so this product is the current that flows into branch 1, measured in amperes. - $I_1 = I_{\text{total}} \frac{R_2}{R_1 + R_2}$: The current through branch 1 equals the total current multiplied by R2 divided by the sum of R1 and R2. Use it when two resistors share one current, and the answer is in the same units as the total current. ## What the formula is doing, and why A **current divider** is a pair of resistors wired in parallel, with one current arriving from a single source. The total current reaches a junction and leaves through two separate branches. The formula answers one question: how much of the total goes through branch 1? The answer is a current in amperes, measured in the same units as the total. **Why the same voltage?** Both resistors connect to the same two points, so the voltage across each one is the same. Call that shared voltage $V$. Ohm's law gives the current in each branch as $I = V / R$, so the current in branch 1 is $V / R_1$ and the current in branch 2 is $V / R_2$. A smaller resistance gives a bigger current, because the same voltage drives more current through the easier path. **Why does $R_2$ go on top?** The branch currents must add up to the total, so $\frac{V}{R_1} + \frac{V}{R_2} = I_{\text{total}}$. Solving for $V$ gives $V = I_{\text{total}} \frac{R_1 R_2}{R_1 + R_2}$. Put that back into $I_1 = \frac{V}{R_1}$ and the $R_1$ on the bottom cancels against the $R_1$ on top, leaving $I_1 = I_{\text{total}} \frac{R_2}{R_1 + R_2}$. The result shows that a bigger $R_2$ leaves less reason to use branch 2, so more current goes into branch 1. **When the formula does not apply.** The formula is for exactly two branches in parallel. With three or more branches, each branch takes a share equal to one divided by its resistance, divided by the sum of one divided by every resistance. In a **series** connection the same current flows through every resistor, so there is nothing to split, and this formula should not be used. **How to check an answer.** The two branch currents must add up to the total, so $I_1 + I_2 = I_{\text{total}}$. Also, the voltage across each branch is the same, so $I_1 R_1$ must equal $I_2 R_2$. If either check fails, a number or a formula has gone wrong, and it is worth finding the mistake before going further. ## An everyday picture Think of a river that splits around an island. Both channels start at the same point upstream and meet again downstream, so they share the same drop in height. In our circuit, the same voltage across both resistors plays the role of that drop. The wide channel carries more water, the way a low resistance carries more current. The narrow channel carries less, and the split depends on how wide each channel is. The water in the two channels adds up to the river's total flow, in the same way the two branch currents add up to the total current. ## Worked example: Splitting 10 mA between a 100 ohm and a 400 ohm resistor Picture a small learning circuit built on a breadboard, the plastic board used to try out electronics. A total current of 10 mA comes in from the supply and reaches a junction. From the junction, the current can go through two resistors wired in parallel: $R_1 = 100\ \Omega$ and $R_2 = 400\ \Omega$. Find the current in each branch. The numbers are chosen for practice and are not taken from a real circuit. 1. **Convert the total current to amperes.** The formula needs every current in amperes, so we change the milliamperes first. The prefix milli means one thousandth, so we divide by 1000: $10\ \text{mA} = \frac{10}{1000}\ \text{A} = 0.01\ \text{A}$. 2. **Write the formula and see what each part does.** The current in branch 1 is the total current times a share. The share has $R_2$ on top and the total of both resistances on the bottom. So the other resistor, $R_2$, decides how much of the current goes into $R_1$: $I_1 = 0.01\ \text{A} \times \frac{R_2}{R_1 + R_2}$. 3. **Find the current through $R_1$.** Put in $R_2 = 400\ \Omega$ on top and $R_1 + R_2 = 100\ \Omega + 400\ \Omega = 500\ \Omega$ on the bottom. The ohms cancel, because ohms divided by ohms leaves a plain number. So $I_1 = 0.01\ \text{A} \times \frac{400}{500} = 0.01\ \text{A} \times 0.8 = 0.008\ \text{A}$. 4. **Find the current through $R_2$ the same way.** The share for branch 2 has $R_1$ on top, because the resistor in the other branch sets the share. So $I_2 = 0.01\ \text{A} \times \frac{100}{500} = 0.01\ \text{A} \times 0.2 = 0.002\ \text{A}$. The smaller resistor gets the larger share, and the calculation confirms that. 5. **Check that the two currents add up.** Every current that enters the junction must leave it, so the two branches together must carry the full 10 mA. Adding them gives $0.008\ \text{A} + 0.002\ \text{A} = 0.010\ \text{A}$, which is the total we started with. If this sum were different, one of the branch currents would be wrong. 6. **Check the voltage across the pair.** Both branches share one voltage. Two resistors in parallel act like one resistor whose value is their product divided by their sum: $\frac{100 \times 400}{100 + 400} = \frac{40\,000}{500} = 80\ \Omega$. The shared voltage is the total current times this value: $0.01\ \text{A} \times 80\ \Omega = 0.8\ \text{V}$. Each branch must match it, so $0.8\ \text{V} / 100\ \Omega = 0.008\ \text{A}$ and $0.8\ \text{V} / 400\ \Omega = 0.002\ \text{A}$, the same as before. 7. **Compare the two branches.** The ratio of the two currents is $\frac{0.008}{0.002} = 4$. The ratio of the two resistances is $\frac{400}{100} = 4$. The two ratios are the same size but point in opposite directions: the branch with four times less resistance carries four times more current. That is the pattern the formula predicts for any pair of parallel resistors. **Answer:** The 100 ohm branch carries **8 mA** (0.008 A) and the 400 ohm branch carries **2 mA** (0.002 A). The smaller resistor takes four times the current, and the two branches add up to the 10 mA that came in. ## Common mistakes - Swapping the resistors in the formula. The current through $R_1$ has $R_2$ on top, so a bigger $R_2$ sends more current into $R_1$. Check the placement by asking which branch should get the bigger share: the one with the smaller resistance. - Using the current divider for resistors in series. In series the same current flows through both resistors, so there is nothing to split. The formula only works when the resistors connect to the same two points. - Using one branch's current as the starting value. The formula needs the total current that enters the junction. The branch current is the answer you are looking for, not the input. - Forgetting the unit prefix. Milliamperes must be converted to amperes before you multiply, or the answer is 1000 times too big. Ten milliamperes is 0.01 amperes, not 10 amperes. ## Where you will meet it - Shunt resistors in ammeters. A shunt is placed in parallel with the sensitive meter part, so most of the current goes around it. - Estimating how one supply current divides between two loads connected in parallel, such as two lamps on the same pair of wires. - Setting the working currents in transistor circuits, where a pair of resistors often shares a current between two branches. - Checking whether a parallel branch will overheat. The current it carries sets how much heat its resistor produces, so the split tells you where the heat goes. ## Questions people ask ### Why does the smaller resistor get more current? Both branches have the same voltage across them. Ohm's law says current equals voltage divided by resistance, so for the same voltage a smaller resistance gives a bigger current. Think of the easier path as the one that takes more of the traffic. ### Is the current divider the same as the voltage divider? They are mirror images. A **voltage divider** shares a voltage between resistors in series, so the same current flows through all of them. A **current divider** shares a current between resistors in parallel, so the same voltage appears across all of them. ### Can I use the formula with three resistors? Not in this two-resistor form. For more branches, each branch takes a share equal to one divided by its resistance, divided by the sum of one divided by every resistance. For two resistors, that rule gives exactly the same answer as the formula on this page. ### What if one branch is a short circuit? A short circuit is a path with zero resistance, so it takes all the current. Put $R_1 = 0$ into the formula and the share for branch 1 becomes $\frac{R_2}{0 + R_2} = 1$. So all the current goes into branch 1, and branch 2 gets none. ## Related formulas - [Voltage Divider](https://formulabro.yunusemrevurgun.com/electricity/voltage-divider/) - [Ohm's Law](https://formulabro.yunusemrevurgun.com/electricity/ohms-law/) - [Resistors in Series and Parallel](https://formulabro.yunusemrevurgun.com/electricity/resistors-in-series-and-parallel/) - [Kirchhoff's Current Law](https://formulabro.yunusemrevurgun.com/electricity/kirchhoffs-current-law/) ## Sources - [Current divider, Wikipedia](https://en.wikipedia.org/wiki/Current_divider) - [Ammeter, Wikipedia](https://en.wikipedia.org/wiki/Ammeter) - [Series and parallel circuits, Wikipedia](https://en.wikipedia.org/wiki/Series_and_parallel_circuits) From Formula Bro (https://formulabro.yunusemrevurgun.com/), a side project developed by yunusemrevurgun.com.