Lesson 1.2 · Module 1: DC Physics & Network Theorems

Kirchhoff's Current & Voltage Laws Explained

Two batteries feeding one shared resistor. Real circuits are networks like this, and two simple laws let you work out every current in them.

The live, interactive simulation for this lesson opens here when JavaScript is on.

Key ideas

At any junction, the current flowing in equals the current flowing out (Kirchhoff’s current law).
Going all the way round any loop, the voltage rises and drops add up to zero (Kirchhoff’s voltage law).
A node is one connected set of wires, all at the same voltage. Node A here is where the three resistors meet.

Try this in the simulation

Hover the wires around Node A: the currents I1 and I2 coming in add up to I3 going down.
Change V2 until its current becomes zero. Then V2 is “just watching” and not supplying anything.
Look at the KCL readout box: it always shows 0 A — charge never piles up.

What are Kirchhoff’s current and voltage laws?

Kirchhoff’s current law says that the currents flowing into any junction add up to the currents flowing out, because charge cannot pile up. Kirchhoff’s voltage law says that if you walk once round any closed loop, the voltage rises and drops add up to zero. Together the two laws let you work out every current and voltage in a circuit that has more than one battery or loop.

What you’ll learn

What a node is, and how to apply the current law at it.
How to walk round a loop and apply the voltage law.
How to write the two laws as equations for a circuit with two batteries.
How to check your answer: the live readout shows the currents at a node adding up to zero.

The formula, letter by letter

\begin{aligned}\text{KCL:}\;& \htmlClass{mv mv-i1}{\htmlStyle{font-size:1.21em}{\htmlData{sym=I_1}{I_1}}} + \htmlClass{mv mv-i2}{\htmlStyle{font-size:1.08em}{\htmlData{sym=I_2}{I_2}}} = \htmlClass{mv mv-i3}{\htmlStyle{font-size:1.21em}{\htmlData{sym=I_3}{I_3}}}\\ & \htmlData{funit=mA,v=0.006185185185185184}{\text{6.19 mA}} + \htmlData{funit=µA,v=-0.0003703703703703708}{\text{-370 }\mu\text{A}} = \htmlData{funit=mA,v=0.005814814814814816}{\text{5.81 mA}}\\ \text{KVL:}\;& \htmlData{sym=U_1}{U_1} - \htmlData{sym=I_1}{I_1}\htmlData{sym=R_1}{R_1} - \htmlClass{mv mv-v_a}{\htmlStyle{font-size:1.00em}{\htmlData{sym=U_A}{U_{\mathrm{A}}}}} = 0\\ \htmlClass{mv mv-v_a}{\htmlStyle{font-size:1.00em}{\htmlData{sym=U_A}{U_{\mathrm{A}}}}} &= \frac{\htmlData{sym=U_1}{U_1}/\htmlData{sym=R_1}{R_1} + \htmlData{sym=U_2}{U_2}/\htmlData{sym=R_2}{R_2}}{1/\htmlData{sym=R_1}{R_1}+1/\htmlData{sym=R_2}{R_2}+1/\htmlData{sym=R_3}{R_3}} = \htmlData{funit=V,v=5.814814814814816}{\text{5.81 V}}\end{aligned}

Shown with this lesson’s starting values. In the simulation the numbers change as you move the sliders.

$I_1$: Current in branch 1 (A · ampere) : How much charge flows each second.
$I_2$: Current in branch 2 (A · ampere) : How much charge flows each second.
$I_3$: Current in branch 3 (A · ampere) : How much charge flows each second.
$U_1$: Voltage of source V1 (V · volt) : The electrical push between two points.
$R_1$: Resistance of R1 (Ω · ohm) : How strongly something opposes current.
$U_{\mathrm{A}}$: Voltage at node A (V · volt) : Measured from node A to ground.
$U_2$: Voltage of source V2 (V · volt) : The electrical push between two points.
$R_2$: Resistance of R2 (Ω · ohm) : How strongly something opposes current.
$R_3$: Resistance of R3 (Ω · ohm) : How strongly something opposes current.

Common questions

What is Kirchhoff’s current law in simple words?

Whatever current flows into a junction must flow out again. If 3 A comes in on one wire, the wires leaving it carry 3 A between them.

What is Kirchhoff’s voltage law in simple words?

Go round a loop and add up every voltage rise (a battery) and every voltage drop (a resistor). You end up exactly where you started, so the total is zero.

Can a current come out negative?

Yes. A negative answer only means the current really flows the opposite way to the arrow you drew. The laws still hold.