Kirchhoff's Laws for current and voltage lie at the heart of circuit analysis. With these two laws, plus the equations for individual component (resistor, capacitor, inductor), we have the basic tool set we need to start analyzing circuits.

This article assumes you are familiar with the definitions of node, distributed node, branch, and loop.

You may want to have a pencil and paper nearby to work the example problems.

## Currents into a node

Try to reason through this example by yourself, before we talk about the theory. The schematic below shows four branch currents flowing in and out of a distributed node. The various currents are in milliamps, $\text{mA}$mAstart text, m, A, end text. One of the currents, $\blueD i$istart color #11accd, i, end color #11accd, is not known.

**Problem 1: What is $i$ii? **

[Show Answer]

Here's another example, this time with variable names instead of numerical values. This node happens to have $5$55 branches. Each branch might (or might not) carry a current, labeled $i_1 \,\text{to} \, i_5$i1toi5i, start subscript, 1, end subscript, start text, t, o, end text, i, start subscript, 5, end subscript.

All the arrows are drawn pointing in. This choice of direction is arbitrary. Arrows pointing inward is as good a choice as any at this point. The arrows establish a reference direction for what we choose to call a positive current.

Look at branch current ${i_1}$i1i, start subscript, 1, end subscript.**Where does it go?**

The first thing ${i_1}$i1i, start subscript, 1, end subscript does is flow into the node (represented by the black dot).

**Then what?**

Here's two things ${i_1}$i1i, start subscript, 1, end subscript can't do: The flowing charge in ${i_1}$i1i, start subscript, 1, end subscript can't stay inside the node. (The node does not have a place to store charge). And ${i_1}$i1i, start subscript, 1, end subscript's charge can't jump off the wires into thin air. Charge just doesn't do that under normal circumstances.

What's left?: The current *has* to flow out of the node through one or more of the other branches.

For our example node, we would write this as,

$i_1 + i_2 + i_3 + i_4 + i_5 = 0$i1+i2+i3+i4+i5=0i, start subscript, 1, end subscript, plus, i, start subscript, 2, end subscript, plus, i, start subscript, 3, end subscript, plus, i, start subscript, 4, end subscript, plus, i, start subscript, 5, end subscript, equals, 0

If ${i_1}$i1i, start subscript, 1, end subscript is a positive current flowing into the node, then one or more of the other currents must be flowing out. Those outgoing currents will have a $-$−minusnegative sign.

This observation about currents flowing in a node is nicely captured in general form as Kirchhoff's Current Law.

## Kirchhoff's Current Law

Kirchhoff's Current Law says that the sum of all currents flowing into a node equals the sum of currents flowing out of the node. It can be written as,

$\large \displaystyle \sum i_{in} = \sum i_{out}$∑iin=∑ioutsum, i, start subscript, i, n, end subscript, equals, sum, i, start subscript, o, u, t, end subscript

[What is the zig-zag symbol?]

### Kirchhoff's Current Law - concept checks

Currents are in milliamps, $\text{mA}$mAstart text, m, A, end text.

** Problem 2: What is $i_5$i5i, start subscript, 5, end subscript? **

[Show Answer]

**Problem 3: What is $i_3$i3i, start subscript, 3, end subscript in this distributed node? **

[Show Answer]

## Voltage around a loop

Below is a circuit with four resistors and a voltage source. We will solve this from scratch using Ohm's Law. Then we will look at the result and make some observations. The first step in solving the circuit is to compute the current. Then we will compute the voltage across the individual resistors.

We recognize this as a series circuit, so there is only one current flowing, $\blueD i$istart color #11accd, i, end color #11accd, through all five elements. To find $\blueD i$istart color #11accd, i, end color #11accd, the four series resistors can be reduced to a single equivalent resistor:

$R_{series} = 100 + 200 + 300 + 400 = 1000\,\Omega$Rseries=100+200+300+400=1000ΩR, start subscript, s, e, r, i, e, s, end subscript, equals, 100, plus, 200, plus, 300, plus, 400, equals, 1000, \Omega

Using Ohm's Law, the current is:

$\blueD i = \dfrac{V}{R_{series}} = \dfrac{20\,\text V}{1000\,\Omega} = 0.020\,\text A = 20 \,\text{mA}$i=RseriesV=1000Ω20V=0.020A=20mAstart color #11accd, i, end color #11accd, equals, start fraction, V, divided by, R, start subscript, s, e, r, i, e, s, end subscript, end fraction, equals, start fraction, 20, start text, V, end text, divided by, 1000, \Omega, end fraction, equals, 0, point, 020, start text, A, end text, equals, 20, start text, m, A, end text

Now we know the current. Next we find the voltages across the four resistors. Go back to the original schematic and add voltage labels to all five elements:

Apply Ohm's Law four more times to find the voltage across each resistor:

$v\phantom{_{\text{R1}}} = \blueD i\,\text R$vR1=iRv, empty space, equals, start color #11accd, i, end color #11accd, start text, R, end text

$v_{\text{R1}} = 20\,\text{mA} \cdot 100\,\Omega = +2\,\text{V}$vR1=20mA⋅100Ω=+2Vv, start subscript, start text, R, 1, end text, end subscript, equals, 20, start text, m, A, end text, dot, 100, \Omega, equals, plus, 2, start text, V, end text

$v_{\text{R2}} = 20\,\text{mA} \cdot 200\,\Omega = +4\,\text{V}$vR2=20mA⋅200Ω=+4Vv, start subscript, start text, R, 2, end text, end subscript, equals, 20, start text, m, A, end text, dot, 200, \Omega, equals, plus, 4, start text, V, end text

$v_{\text{R3}} = 20\,\text{mA} \cdot 300\,\Omega = +6\,\text{V}$vR3=20mA⋅300Ω=+6Vv, start subscript, start text, R, 3, end text, end subscript, equals, 20, start text, m, A, end text, dot, 300, \Omega, equals, plus, 6, start text, V, end text

$v_{\text{R4}} = 20\,\text{mA} \cdot 400\,\Omega = +8\,\text{V}$vR4=20mA⋅400Ω=+8Vv, start subscript, start text, R, 4, end text, end subscript, equals, 20, start text, m, A, end text, dot, 400, \Omega, equals, plus, 8, start text, V, end text

We know the current and all voltages. The circuit is now solved.

We can write the voltages for the resistors and the source on the schematic. These five voltages are referred to as *element voltages*. (The circuit nodes get names, $\greenE{\text a}$astart color #0d923f, start text, a, end text, end color #0d923f to $\greenE{\text e}$estart color #0d923f, start text, e, end text, end color #0d923f, so we can talk about them.)

Let's do a quick check. Add up the voltages across the resistors,

$2\,\text{V} + 4\,\text{V} + 6\,\text{V} + 8\,\text{V} = 20 \,\text V$2V+4V+6V+8V=20V2, start text, V, end text, plus, 4, start text, V, end text, plus, 6, start text, V, end text, plus, 8, start text, V, end text, equals, 20, start text, V, end text

The individual resistor voltages add up to the source voltage. This makes sense, and confirms our calculations.

Now we add up the voltages again, using a slightly different procedure: by "going around the loop." There's no new science here, we are just rearranging the same computation.

### Procedure: Add element voltages around a loop

**Step 1:** Pick a starting node.

**Step 2:** Pick a direction to travel around the loop (clockwise or counterclockwise).

**Step 3:** Walk around the loop.

[hint]

Include element voltages in a growing sum according to these rules:

- When you encounter a new element, look at the voltage sign as you enter the element.
- If the sign is $+$+plus, then there will be a voltage
*drop*going through the element.*Subtract*the element voltage. - If the sign is $-$−minus, then there will be a voltage
*rise*going through the element.*Add*the element voltage. [Alternative rule set]

**Step 4:** Continue around the loop until you reach the starting point, including element voltages all the way around.

### Apply the loop procedure

Let's follow the procedure step-by-step.

Start at the lower left at node $\greenE{\text a}$astart color #0d923f, start text, a, end text, end color #0d923f.

Walk clockwise.

A copy of the solved circuit.

- The first element we come to is the voltage source. The first voltage sign we encounter is a $-$−minus minus sign, so there is going to be a voltage
*rise*going through this element. Consulting the procedure step 3., we initialize the loop sum by*adding*the source voltage.

$v_{loop} = +20\,\text V$vloop=+20Vv, start subscript, l, o, o, p, end subscript, equals, plus, 20, start text, V, end text going through the voltage source, to node $\greenE{\text b}$bstart color #0d923f, start text, b, end text, end color #0d923f.

The next element we encounter is the $100\,\Omega$100Ω100, \Omega resistor. Its nearest voltage sign is $+$+plus. Consult the procedure again, and this time we *subtract* the element voltage from the growing sum.

$v_{loop} = + 20\,\text V - 2\,\text V$vloop=+20V−2Vv, start subscript, l, o, o, p, end subscript, equals, plus, 20, start text, V, end text, minus, 2, start text, V, end text going through the $100\,\Omega$100Ω100, \Omega resistor, to node $\greenE{\text c}$cstart color #0d923f, start text, c, end text, end color #0d923f.

Keep going. Next we visit the $200\,\Omega$200Ω200, \Omega resistor, and again we first encounter a $+$+plus sign, so we subtract this voltage.

$v_{loop} = + 20\,\text V - 2\,\text V - 4\,\text V$vloop=+20V−2V−4Vv, start subscript, l, o, o, p, end subscript, equals, plus, 20, start text, V, end text, minus, 2, start text, V, end text, minus, 4, start text, V, end text going through the $200\,\Omega$200Ω200, \Omega resistor, to node $\greenE{\text d}$dstart color #0d923f, start text, d, end text, end color #0d923f.

We complete the loop with the addition of two more elements,

$v_{loop} = + 20\,\text V - 2\,\text V - 4\,\text V - 6\,\text V \,$vloop=+20V−2V−4V−6Vv, start subscript, l, o, o, p, end subscript, equals, plus, 20, start text, V, end text, minus, 2, start text, V, end text, minus, 4, start text, V, end text, minus, 6, start text, V, end text through the $300\,\Omega$300Ω300, \Omega resistor, to node $\greenE{\text e}$estart color #0d923f, start text, e, end text, end color #0d923f.

$v_{loop} = + 20\,\text V - 2\,\text V - 4\,\text V - 6\,\text V - 8\,\text V\,$vloop=+20V−2V−4V−6V−8Vv, start subscript, l, o, o, p, end subscript, equals, plus, 20, start text, V, end text, minus, 2, start text, V, end text, minus, 4, start text, V, end text, minus, 6, start text, V, end text, minus, 8, start text, V, end text after the $400\,\Omega$400Ω400, \Omega resistor.

(Check the circuit diagram, make sure I got the last two $-$−minus signs correct.)

- Done. We made it back home to node $\greenE{\text a}$astart color #0d923f, start text, a, end text, end color #0d923f. What does this expression for $v_{loop}$vloopv, start subscript, l, o, o, p, end subscript add up to?

$v_{loop} = + 20\,\text V - 2\,\text V - 4\,\text V - 6\,\text V - 8\,\text V = 0$vloop=+20V−2V−4V−6V−8V=0v, start subscript, l, o, o, p, end subscript, equals, plus, 20, start text, V, end text, minus, 2, start text, V, end text, minus, 4, start text, V, end text, minus, 6, start text, V, end text, minus, 8, start text, V, end text, equals, 0

The sum of voltages going around the loop is $0$00. The starting and ending node is the same, so the starting and ending voltage is the same. On your "walk" you went up voltage rises and down voltage drops, and they all cancel out when you get back to where you started. This happens because electric force is *conservative*. There isn't a net gain or loss of energy if you return to the same place you started.

We'll do another example, this time with variable names instead of numerical values. The following familiar schematic is labeled with voltages and node names. The voltage polarity on the resistors is arranged in a way you might not expect, with all the arrows pointing in the same direction around the loop. This reveals a cool property of loops.

Let's take a walk around the loop, adding up voltages as we go. Our starting point is node $\greenE{\text a}$astart color #0d923f, start text, a, end text, end color #0d923f in the lower left corner. Our walk goes clockwise around the loop (an arbitrary choice, either way works).

Starting at node $\greenE{\text a}$astart color #0d923f, start text, a, end text, end color #0d923f, going up, we first encounter a minus sign on the voltage source, which says there is going to be a voltage *rise* of $v_{ab}$vabv, start subscript, a, b, end subscript volts going through the voltage source. Because it is a voltage rise, this element voltage gets a $+$+plus sign when we include it in the loop sum.

Continue around the loop from node $\greenE{\text b}$bstart color #0d923f, start text, b, end text, end color #0d923f to $\greenE{\text c}$cstart color #0d923f, start text, c, end text, end color #0d923f to $\greenE{\text d}$dstart color #0d923f, start text, d, end text, end color #0d923f to $\greenE{\text e}$estart color #0d923f, start text, e, end text, end color #0d923f, and finish back home at node $\greenE{\text a}$astart color #0d923f, start text, a, end text, end color #0d923f. Append resistor voltages to the loop sum along the way. The polarity labels on all the resistors are arranged so we encounter a $-$−minus sign as we approach each resistor. So the resistor voltages all go into the loop sum with a $+$+plus sign. The final loop sum looks like this:

$+v_{\text{ab}} + v_{\text{R1}} + v_{\text{R2}} + v_{\text{R3}} + v_{\text{R4}}$+vab+vR1+vR2+vR3+vR4plus, v, start subscript, start text, a, b, end text, end subscript, plus, v, start subscript, start text, R, 1, end text, end subscript, plus, v, start subscript, start text, R, 2, end text, end subscript, plus, v, start subscript, start text, R, 3, end text, end subscript, plus, v, start subscript, start text, R, 4, end text, end subscript

What does this add up to? Let's reason it out.

The loop starts and ends at the same node, so the starting and ending voltages are identical. We went around the loop, adding voltages, and we end up back at the same voltage. That means the voltages have to add to zero. For our example loop, we would write this as,

$v_{\text{ab}} + v_{\text{R1}} + v_{\text{R2}} + v_{\text{R3}} + v_{\text{R4}} = 0$vab+vR1+vR2+vR3+vR4=0v, start subscript, start text, a, b, end text, end subscript, plus, v, start subscript, start text, R, 1, end text, end subscript, plus, v, start subscript, start text, R, 2, end text, end subscript, plus, v, start subscript, start text, R, 3, end text, end subscript, plus, v, start subscript, start text, R, 4, end text, end subscript, equals, 0

[What would happen if voltages around a loop didn't add up to zero?]

[What happens if you go around the loop the other way?]

[one hint]

[show answer]

This observation about voltages around a loop is nicely captured in general form as Kirchhoff's Voltage Law.

## Kirchhoff's Voltage Law

Kirchhoff's Voltage Law: *The sum of voltages around a loop is zero.*

Kirchhoff's Voltage Law can be written as,

$\large\displaystyle \sum_n v_n = 0$n∑vn=0sum, start subscript, n, end subscript, v, start subscript, n, end subscript, equals, 0

where $n$nn counts the element voltages around the loop.

You can also state Kirchhoff's Voltage Law another way: *The sum of voltage rises equals the sum of voltage drops around a loop.*

$\large \displaystyle \sum v_{rise} = \sum v_{drop}$∑vrise=∑vdropsum, v, start subscript, r, i, s, e, end subscript, equals, sum, v, start subscript, d, r, o, p, end subscript

Kirchhoff's Voltage Law has some nice properties:

- You can trace a loop starting from
*any*node. Walk around the loop and end up back at the starting node, the sum of voltages around the loop adds up to zero. - You can go around the loop in either direction, clockwise or counterclockwise. Kirchhoff's Voltage Law still holds.
- If a circuit has multiple loops, Kirchhoff's Voltage Law is true for every loop.

#### Voltages all positive?

If you are wondering: how can the element voltages all be positive if they have to add up to zero? It's okay. The voltage arrows and polarity signs are just reference directions for voltage. When the circuit analysis is complete, one or more of the element voltages around the loop will be negative with respect to its voltage arrow. The signs of the actual voltages always sort themselves out during calculations.

### Kirchhoff's Voltage Law - concept check

**Problem 4: What is $v_{R3}$vR3v, start subscript, R, 3, end subscript? **

Reminder: Check the first sign of each element voltage as you walk around the loop.

[Show Answer]

## Summary

We were introduced to two new friends.

Kirchhoff's Current Law for branch currents at a node,

$\large\displaystyle \sum_n i_n = 0$n∑in=0sum, start subscript, n, end subscript, i, start subscript, n, end subscript, equals, 0

Kirchhoff's Voltage Law for element voltages around a loop,

$\large\displaystyle \sum_n v_n = 0$n∑vn=0sum, start subscript, n, end subscript, v, start subscript, n, end subscript, equals, 0

Our new friends sometimes go by their initials, KCL and KVL.

And we learned it's important to pay close attention to voltage and current signs if we want correct answers. This is a tedious process that requires attention to detail. It is a core skill of a good electrical engineer.

## FAQs

### What is Kirchhoff's rule Khan Academy? ›

Kirchhoff's Voltage Law: **The sum of voltages around a loop is zero**. where n counts the element voltages around the loop. You can also state Kirchhoff's Voltage Law another way: The sum of voltage rises equals the sum of voltage drops around a loop.

**What is Kirchhoff's law for dummies? ›**

Kirchhoff's Laws: The Basics

Specifically, the laws state: **The algebraic sum of current into any junction is zero**. Since current is the flow of electrons through a conductor, it cannot build up at a junction, meaning that current is conserved: What goes in must come out.

**What is the law of Kirchhoff's current law? ›**

Kirchhoff's Current Law, often shortened to KCL, states that “**The algebraic sum of all currents entering and exiting a node must equal zero**.” This law is used to describe how a charge enters and leaves a wire junction point or node on a wire.

**What are Kirchhoff's three laws? ›**

Whenever you are studying the light from an astronomical object, recall that there are three things you need to consider: **the emission of the light by the source,** **processes that affect the light during its travel from the source to the observer, and**. **the process of detection of the light by the observer**.

**How many Kirchhoff's laws are there? ›**

Kirchhoff's circuit laws are two equalities that deal with the current and potential difference (commonly known as voltage) in the lumped element model of electrical circuits. They were first described in 1845 by German physicist Gustav Kirchhoff.

**What are the two types of Kirchhoff's law? ›**

**Kirchhoff's first law is based on the conservation of charges, and Kirchhoff's second law is based on energy conservation**. Gustav Robert Kirchhoff described it in 1845. Kirchhoff's first law is junction rule or current law, and Kirchhoff's second law is loop rule or voltage law.

**What is Kirchhoff's 1st and 2nd law? ›**

Kirchhoff's first law is based on the conservation of charge because sum of current entering to the junction is equal to sum of current leaving the junction. Kirchhoff's second law states that the algebraic sum of potential drops in a closed circuit is zero.

**What is the basic fact of the Kirchhoff's first law? ›**

The law states that at any circuit junction, the sum of the currents flowing into and out of that junction are equal. In simple terms, what KCL really says is that, **The sum of all currents entering a node is equal to the sum of all currents leaving the node**.

**What is the significance of Kirchhoff's law? ›**

Kirchhoff's First Law Overview

It is a direct application of the electric charge conservation principle. The law simply states that the sum of the currents flowing out of the junction is equal in value with the sum of currents flowing out of that junction. The junction can be any node present inside the circuit.

**What is Kirchhoff's law 2nd law? ›**

Kirchhoff's voltage law (2nd Law) states that in any complete loop within a circuit, the sum of all voltages across components which supply electrical energy (such as cells or generators) must equal the sum of all voltages across the other components in the same loop.

### What is Kirchhoff's law example? ›

Kirchhoff's Current Law Example No2

**At point A, I _{1} is equal to I_{T}, thus there will be an I_{1}*R voltage drop across resistor R_{1}**. The circuit has 2 branches, 3 nodes (B, C and D) and 2 independent loops, thus the I*R voltage drops around the two loops will be: Loop ABC ⇒ 12 = 4I

_{1}+ 6I. Loop ABD ⇒ 12 = 4I

_{1}+ 12I.

**What is Kirchhoff's law conclusion? ›**

Kirchhoff's Law states that the emissivity of a surface is equal to its absorbance, where the absorbance of a surface is the ratio of the radiant power absorbed to the radiant power incident on the surface. It concludes that good radiators of thermal radiations are good absorbers.

**Which circuit can only be solved by Kirchhoff's law? ›**

Kirchhoffs Current Law

Also for current to flow either in or out of a node a closed circuit path must exist. We can use Kirchhoff's current law when analysing parallel circuits.

**What is the difference between Kirchhoff's voltage law and Kirchhoff's current law? ›**

WHAT IS KVL AND KCL? Kirchhoff's Voltage Law (KVL) Kirchhoff's voltage law states that the algebraic sum of the voltages around any loops in a circuit is always zero. Kirchhoff's Current Law (KCL) Kirchhoff current law states that the algebraic sum of all currents entering a node of a circuit is always zero.

**What is the another name of Kirchhoff's second law? ›**

Kirchhoff's second law, also known as **Kirchhoff's voltage law** (KVL) states that the sum of all voltages around a closed loop in any circuit must be equal to zero. This again is a consequence of charge conservation and also conservation of energy.

**Why do we use Kirchhoff's law instead of Ohm's law? ›**

Answer and Explanation: Kirchhoff's rule is more often preferred to analyze a circuit because **it doesn't have limitations like other circuit analyzing techniques have**. Ohm's law is used in Kirchhoff's rules.

**What are two applications of Kirchhoff's law in daily life? ›**

1) **Sand is rough black, so it is a good absorber and hence in deserts, days will be very hot**. Now in accordance with Kirchhoff's law, good absorber is a good emitter so nights will be cold. This is why days are hot and nights are cold in desert. 2) Sodium vapors, on heating, emit two bright yellow lines.

**Why is Kirchhoff's law better than Ohm's law? ›**

While Ohm's Law is applicable to a resistive element, **Kirchhoff's Laws are applied to a series of elements**. This is the most important difference between Ohm's law and Kirchhoff's law. KCL and KVL are usually used in circuit analysis together with Ohm's Law.

**What is Kirchhoff's first current law 1 state? ›**

Kirchoff's voltage law states that the algebraic sum of the product of resistance and current in each part of any closed circuit is equal to the algebraic sum of the emf's in that closed circuit.

**What is another term of Kirchhoff's first law? ›**

Kirchhoff's first law of electrical circuits is based on charge conservation, while Kirchhoff's second law of electrical circuits is based on energy conservation. Kirchhoff's concept is known as the Conservation of Charge because the current is preserved around the junction with no current loss.

### Is Kirchhoff's law accurate? ›

With the exception of the voltage law applied to high resistance circuits, we conclude that **Kirchhoff's Laws accurately predict the behavior of resistive circuits**. (KCL). analysis and design of electrical circuits.

**What's true about Kirchhoff's rules? ›**

Kirchhoff's loop rule states that **the algebraic sum of potential differences, including voltage supplied by the voltage sources and resistive elements, in any loop must be equal to zero**. For example, consider a simple loop with no junctions, as in Figure 6.3.

**What are Kirchhoff's rules used for? ›**

Kirchhoff's laws, one for voltage and one for current, **determine what a connection between circuit elements means**. These laws can help us analyze this circuit. The places where circuit elements attach to each other are called. At every node, the sum of all currents entering a node must equal zero.

**What is Kirchhoff's law quizlet? ›**

Kirchhoff's First Law. **The sum of the currents entering any point (or junction) in a circuit is equal to the sum of the currents leaving that same point**.

**Why we use Kirchhoff's rule? ›**

Kirchhoff's law is used **to find:** **The values of current, voltage and internal resistance in DC circuits**. By applying this law we can also find the unknown resistance in the circuit. Wheatstone bridge is an important application of Kirchhoff's law.

**What is Kirchhoff's first law called? ›**

Kirchhoff's first law goes by several names as **Kirchhoff's Current Law (KCL), Kirchhoff's Junction Rule, Kirchhoff's point rule, Kirchhoff's nodal rule**. It is an application of the principle of conservation of electric charge.

**Do Kirchhoff's rules always apply? ›**

Kirchhoff's rules can be applied to any circuit, regardless of its composition and structure. Because combining elements is often easy in parallel and series, **it is not always convenient to apply Kirchhoff's rules**. To solve for current in a circuit, the loop and junction rules can be applied.