Kirchhoff's Current Law

In one sentence: Kirchhoff’s current law says that the electricity flowing into a junction has to flow out again, because nothing piles up or disappears at a junction.

$\sum I_{\text{in}} = \sum I_{\text{out}}$

Read it out loud: “The sum of the currents into a junction equals the sum of the currents out of it. The sum sign means add up every term in the list. I in is the current flowing into the junction, and I out is the current flowing out of it, both measured in amperes.”

Before you start

The ideas and signs this page uses, explained from zero. For powers of ten, Greek letters and other notation in general, see Math basics.

The sum sign $\sum$: The capital Greek letter sigma, $\sum$ , is the sum sign. It means ‘add up every term in the list’. So $\sum I_{\text{in}}$ means add all the incoming currents together. For example, if three wires bring in 2 A, 3 A and 1 A, then $\sum I_{\text{in}} = 2 + 3 + 1 = 6$ A. The sum sign does not change the numbers, it only tells you to add them all up.
Subscripts such as in and out: A subscript is a small label written under a letter. $I_{\text{in}}$ is read ‘I in’ and means the current flowing in. $I_{\text{out}}$ is read ‘I out’ and means the current flowing out. The letter $I$ is the same kind of quantity in both, and the subscript only tells you which direction it flows, relative to the junction.
Electric current and the ampere (A): Electric current, written $I$ , is the flow of electric charge through a wire. Its unit is the ampere, written A, and one ampere means one coulomb of charge passes a point every second. A coulomb is a unit of charge, about $6.24 \times 10^{18}$ electrons. A current of 5 A means 5 coulombs pass a point each second.
Junction, also called a node: A junction, also called a node, is a point where three or more wires meet. Two wires joined end to end is not usually called a junction, because nothing else branches off. At a junction current can split into several branches or join from several wires, and the rule on this page says what the total in and out must match.
Powers of ten, like $10^{18}$: A power of ten is 10 multiplied by itself a set number of times, written with a small raised number called the exponent. $10^{3}$ means $10 \times 10 \times 10 = 1000$ , and $10^{18}$ means 10 multiplied by itself 18 times, which is 1 followed by 18 zeros. So $6.24 \times 10^{18}$ means 6.24 with the decimal point moved 18 places to the right, about 6.24 billion billion.
Conservation of charge: Conservation means a quantity is never created or destroyed, only moved from place to place. Electric charge is conserved, which means charge does not appear or vanish at a junction. The current rule is charge conservation written for every junction, and it is one of the most reliable rules in physics.

What each symbol means

Symbol	Meaning
$\sum$	The sum sign, read ‘the sum of’. It tells you to add up every term of the kind that follows it, for every wire that touches the junction. Without it, you would have to write each current out in full, one plus after another.
$I_{\text{in}}$	A current flowing into the junction, measured in amperes (A). One ampere means one coulomb of charge per second. In a junction with two wires coming in, you would add two of these together, each one measured in amperes.
$I_{\text{out}}$	A current flowing out of the junction, measured in amperes (A). One ampere means one coulomb of charge per second. A wire that carries current away from the junction contributes its current to this sum.
$I$	Any current in a wire, which is the flow of charge through that wire, measured in amperes (A). One ampere is one coulomb of charge per second. On a circuit diagram, the arrow beside each wire shows the direction you have chosen for this flow.

The formula piece by piece

$I$: Start with one current. $I$ is the flow of charge through a single wire, in amperes. It counts how many coulombs pass a point each second, and it is the quantity the rule adds up.
$I_{\text{in}}$: A current that arrives at the junction. It is a flow of charge heading in, and it is one of the terms on the left side of the balance. Each wire that brings current in gives one term.
$I_{\text{out}}$: A current that leaves the junction. It is one of the terms on the right side of the balance. Each wire that takes current away gives one term.
$\sum I_{\text{in}}$: Add up the incoming currents. The sum sign tells you to add every term, so two wires bringing in 5 A and 3 A give a total of 8 A coming into the junction.
$\sum I_{\text{in}} = \sum I_{\text{out}}$: The whole formula in one sentence: the total current flowing into a junction equals the total current flowing out of it, because charge cannot pile up at the junction.
$\sum I = 0$: A standard second form. Give each current a sign, positive for currents in and negative for currents out, and add them all. Then the total is zero, which is the same balance written with one side of the equation empty.

What the formula is doing, and why

A junction, also called a node, is a point where three or more wires meet. Charge does not pile up at a junction, so whatever flows in must flow out again. Think of the current in each wire as a steady stream of charge. If more charge came into the junction than left it, the extra would have to stay there and build up, and that does not happen in a wire. This is the conservation of charge, and Kirchhoff’s current law, often shortened to KCL, applies it to every junction.

Current $I$ is the flow of charge, in amperes (A). Suppose 5 A flows into a junction through one wire, and 2 A leaves through a second wire. The third wire must take away the remaining 3 A. The formula writes this balance as the sum of the currents in, $\sum I_{\text{in}}$ , equal to the sum of the currents out, $\sum I_{\text{out}}$ . Why add? Each wire carries its own stream of charge, and the streams merge or split at the junction. The total of the streams in must equal the total of the streams out, Water flowing into a pipe junction has to leave through the other pipes.

Many people prefer one sign convention. Choose a direction for each current on the diagram. Count the currents going into the junction as positive, and count the currents going out as negative. Then the total is zero: $\sum I = 0$ . Both versions give the same answer, so use whichever is easier to read. A negative answer for a current does not mean something is wrong. It means that current actually flows opposite to the direction you chose.

KCL does not say how the current divides between the branches. For that you also need the resistance of each branch and Ohm’s law, which says the voltage across each branch equals its current times its resistance. KCL only says that the total must balance at every junction. Most circuit problems use it together with Ohm’s law and the voltage law.

An everyday picture

Think of a road junction. If 500 cars per hour come in from the north road, then the cars leaving by the east and south roads must also add up to 500 cars per hour, if no car parks or disappears on the way. The cars do not pile up at the junction, and the junction does not use them up. A tee in a water pipe works the same way: the water that enters the tee has to leave through its outlets, and the total leaving is the total arriving.

Worked example: A junction with 5 A in, then a three-way split

There are two junctions. At the first, 5 A flows in on one wire, 2 A leaves on a second wire, and an unknown current $I_3$ leaves on a third wire. At the second, 10 A flows in on one wire and splits into three branches. Those branches carry 3 A and 4 A, and the third branch carries the unknown current $I_3$ . Find the unknown current at each junction.

Step 1: write the balance at the first junction. The current coming in is 5 A. The two wires going out carry 2 A and $I_3$ . So the balance is $\sum I_{\text{in}} = \sum I_{\text{out}}$ , which gives $5\ \text{A} = 2\ \text{A} + I_3$ .
Step 1: solve for the unknown. To leave $I_3$ alone, subtract 2 A from both sides. That gives $I_3 = 5\ \text{A} - 2\ \text{A} = 3\ \text{A}$ . The unknown wire carries 3 A out of the junction.
Step 2: write the balance at the second junction. The current coming in is 10 A. The three wires going out carry 3 A, 4 A and $I_3$ . So the balance is $10\ \text{A} = 3\ \text{A} + 4\ \text{A} + I_3$ . Every wire touching the junction has to be counted, and this junction has three outgoing wires.
Step 2: solve for the unknown. Subtract the known outgoing currents from the incoming total: $I_3 = 10\ \text{A} - 3\ \text{A} - 4\ \text{A} = 3\ \text{A}$ . Subtracting 3 A and then 4 A is the same as subtracting 7 A at once, which leaves 3 A.
Check the total at the second junction. The three outgoing currents add up to $3\ \text{A} + 4\ \text{A} + 3\ \text{A} = 10\ \text{A}$ . That equals the 10 A that came in. If the total did not match, a value or a wire would have been left out, so a matching total is a good sign.
Check the unit. Every term in the balance is in amperes, which means coulombs per second. Adding and subtracting amperes gives amperes, so the balance has the right unit. No powers of ten were needed in this example, because all the currents are whole numbers of amperes.

Answer: At the first junction the unknown wire carries 3 A. At the second junction the third branch also carries 3 A, so the three branches share the 10 A exactly. Charge is conserved at both junctions.

The numbers in this example are recomputed by script every time the site is built.

Common mistakes

Forgetting a wire. A junction with three branches has three currents, not two. Count every wire that touches the junction before you write the balance.
Thinking the current gets used up at a junction. Charge is not consumed there. The total current in always equals the total current out.
Using KCL to find voltages or resistances. KCL only balances currents at a junction. Use Ohm’s law for the voltage across a part.
Assigning a current direction and then forgetting it. A negative answer means the current flows opposite to the direction you chose, so keep the signs consistent.

Where you will meet it

Circuit analysis of printed circuit boards, where every junction has to balance its currents.
Electrical power grids, where the currents at each substation must match the flows on the lines that connect to it.
Nodal analysis, a standard method for solving circuits with many branches, which is built on KCL.
Water and gas pipe networks, which use the same conservation idea for the flow at each junction.

Questions people ask

Does current get used up at a junction?

No. The current leaving a junction equals the current entering it. What the circuit does with the energy is a different question, and it is answered by the resistors and the voltages.

What is a node in a circuit?

A node is a point where wires meet, which is the same thing as a junction. Some textbooks use node for any connection point, and KCL is written for each node.

Is Kirchhoff's current law the same as Ohm's law?

No. Ohm’s law links voltage, current and resistance inside one part. Kirchhoff’s current law balances the current at a junction. Most circuit problems use both rules.

Why do some currents come out negative?

If you use signs, you pick a direction for each current first. A current that comes out negative flows opposite to the direction you picked.

Ohm's Law: Ohm’s law says the push on a wire, called voltage, equals the flow through it, called current, times how hard the wire resists that flow, called resistance.
Resistors in Series and Parallel: Resistors in a line add up, and resistors side by side always give a total smaller than the smallest one.
RC Time Constant: The RC time constant tells you how long a capacitor takes to fill up through a resistor. You get it by multiplying the resistance in ohms by the capacitance in farads, and the answer comes out in seconds.