# Kirchhoff's laws (article) | Circuits | Khan Academy (2023)

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?

• $-4\,\text{mA}$4mAminus, 4, start text, m, A, end text

• $+2\,\text{mA}$+2mAplus, 2, start text, m, A, end text

• $+4\,\text{mA}$+4mAplus, 4, start text, m, A, end text

• $+8\,\text{mA}$+8mAplus, 8, start text, m, A, end text

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

(Video) Kirchhoff's voltage law | Circuit analysis | Electrical engineering | Khan Academy

[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?

• $-6\,\text{mA}$6mAminus, 6, start text, m, A, end text

• $\phantom{+}0\,\text{mA}$+0mAempty space, 0, start text, m, A, end text

• $+6\,\text{mA}$+6mAplus, 6, start text, m, A, end text

• $+9\,\text{mA}$+9mAplus, 9, start text, m, A, end text

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

• $-6\,\text{mA}$6mAminus, 6, start text, m, A, end text

• $\phantom{+}0\,\text{mA}$+0mAempty space, 0, start text, m, A, end text

(Video) Kirchhoff's law application: 2-loop circuit solving | Electric current | Physics | Khan Academy

• $+6\,\text{mA}$+6mAplus, 6, start text, m, A, end text

• $+12\,\text{mA}$+12mAplus, 12, start text, m, A, end text

## 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=20mA100Ω=+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=20mA200Ω=+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=20mA300Ω=+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=20mA400Ω=+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.

(Video) Kirchhoff's Law, Junction & Loop Rule, Ohm's Law - KCl & KVl Circuit Analysis - Physics

### Apply the loop procedure

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

2. Walk clockwise.

A copy of the solved circuit.

1. 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=+20V2Vv, 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=+20V2V4Vv, 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=+20V2V4V6Vv, 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=+20V2V4V6V8Vv, 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.)

1. 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=+20V2V4V6V8V=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]

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,

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

(Video) Kirchhoff's current law | Circuit analysis | Electrical engineering | Khan Academy

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.

• $+24\,\text V$+24Vplus, 24, start text, V, end text

• $+8\,\text V$+8Vplus, 8, start text, V, end text

• $+7\,\text V$+7Vplus, 7, start text, V, end text

• $+6\,\text V$+6Vplus, 6, start text, V, end text

## Summary

We were introduced to two new friends.

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

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

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

$\large\displaystyle \sum_n v_n = 0$nvn=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 the Kirchoff junction rule Khan Academy? ›

Junctions can't store current, and current can't just disappear into thin air because charge is conserved. Therefore, the total amount of current flowing through the circuit must be constant.

What are Kirchhoff's laws Openstax? ›

Kirchhoff's first rule—the junction rule. The sum of all currents entering a junction must equal the sum of all currents leaving the junction. Kirchhoff's second rule—the loop rule. The algebraic sum of changes in potential around any closed circuit path (loop) must be zero.

What is the summary of Kirchhoff's law? ›

Kirchhoff's Laws

(1) The sum of the currents flowing at any instant toward a junction in an electrical network is equal to the sum of the currents flowing away from it. (2) The algebraical sum of the e.m.f.s and p.d.s around any closed electrical circuit is zero.

What is the Kirchhoff's law in simple terms? ›

According to Kirchhoff's Voltage Law, The voltage around a loop equals the sum of every voltage drop in the same loop for any closed network and equals zero. Put differently, the algebraic sum of every voltage in the loop has to be equal to zero and this property of Kirchhoff's law is called conservation of energy.

What is Kirchhoff's junction rule state and explain what is its significance? ›

Kirchhoff's Junction Rule

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 Kirchhoff's I junction rule and II loop rule? ›

Kirchhoff's first rule—the junction rule: The sum of all currents entering a junction must equal the sum of all currents leaving the junction. Kirchhoff's second rule—the loop rule: The algebraic sum of changes in potential around any closed circuit path (loop) must be zero.

What is Kirchhoff 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. So, it is based on the conservation of energy.

Why is Kirchhoff's law important? ›

This law is used to calculate the unknown values of current and voltages in the circuit. Kirchhoff's law was the first law that helped the analysis and calculation of complex circuits become manageable and easy. The Wheatstone bridge is an essential application of Kirchhoff's laws.

What are Kirchhoff's 3rd laws? ›

Third Law: A thin cool gas in front of a hotter solid, liquid, or dense-gas background removes the radiation from the background source at special wave lengths. If the resulting radiation were passed through a prism, there would be dark lines superimposed on the continuous band of colors due to the background.

What are the rules of Kirchhoff's law? ›

KIRCHHOFF'S RULES
• Kirchhoff's first rule—the junction rule. The sum of all currents entering a junction must equal the sum of all currents leaving the junction: (6.3.1)
• Kirchhoff's second rule—the loop rule. The algebraic sum of changes in potential around any closed circuit path (loop) must be zero: (6.3.2)

### Are there any limitations of Kirchhoff's laws? ›

The limitation related to Kirchhoff's law is that it is only applicable when there is the absence of fluctuating magnetic field. this means that in the presence of fluctuating magnetic field this law does not apply.

What is Kirchhoff's first law in one sentence what is the equation? ›

Kirchhoff's Law #1 - The sum of the currents entering a node must equal the sum of the currents exiting a node. The first law is the statement of current conservation. For the node on the right, i1=i2+i3. If all currents had been defined as entering the node, then the sum of the currents would be zero.

How Kirchhoff's laws are justified? ›

These two laws are justified by the fact that the first law of currents shows us the conservation of charge in the circuit while the second law of voltages shows the conservation of energy in the circuit.

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.

What are the two Kirchhoff's laws for electrical circuits? ›

Kirchhoff's first rule—the junction rule: The sum of all currents entering a junction must equal the sum of all currents leaving the junction. Kirchhoff's second rule—the loop rule: The algebraic sum of changes in potential around any closed circuit path (loop) must be zero.

What is Kirchhoff's loop rule in real life? ›

Put another way, Kirchhoff's Laws state that the sum of all currents leaving a node in an electrical network always equals zero. These laws are extremely useful in real life because they describe the relation of values of currents that flow through a junction point and voltages in an electrical circuit loop.

What is Kirchhoff's second law in simple words? ›

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.

What laws does Kirchhoff's second law obey? ›

Thus, it is based on the Law of Conservation of Energy.

Why Kirchhoff's law is better than 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.

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 is Kirchhoff's current law formula? ›

Another way of stating Kirchhoff's Current Law is: ∑ini=∑outi. In the example above the three equations would be: i1=i2Node Ai2+i4=i3Node Bi3=i1+i4Node C. In this formulation, we are saying that the sum of the currents into a node equals the sum of the currents out of that node.

Who invented Kirchhoff's law? ›

Gustav Kirchhoff, in full Gustav Robert Kirchhoff, (born March 12, 1824, Königsberg, Prussia [now Kaliningrad, Russia]—died October 17, 1887, Berlin, Germany), German physicist who, with the chemist Robert Bunsen, firmly established the theory of spectrum analysis (a technique for chemical analysis by analyzing the ...

What are the different types of Kirchhoff's law? ›

The two variations of Kirchhoff's circuit laws are Kirchhoff's current law and Kirchhoff's voltage law. The current law states that all current entering a node must equal the current leaving a node. The voltage law states that the total voltage is equal to the sum of the voltage drops in a closed loop.

What is Ohm's law and Kirchhoff's law? ›

The circuit equations can be determined using Ohm's Law, which gives the relationship between voltage and current in a resistor (V=IR), and Kirchhoff's Current and Voltage Laws, which govern the currents entering and exiting a circuit node and the sum of voltages around a circuit loop, respectively.

What is violation of Kirchhoff's law? ›

In the presence of loss, absorptivity and emissivity are unequal, violating Kirchhoff's law. In a stronger modulation regime, we observe the preservation of nonreciprocal scattering and the emergence of Rabi splitting in the forward scattering problem. This results in an even stronger violation of Kirchhoff's law.

What is Kirchhoff's junction law quizlet? ›

What is Kirchhoff's Junction Law? the total current flowing into and out of any junction point is zero at all times.

What is Kirchhoff's rules Kirchhoff's junction rule a statement of? ›

Kirchhoff's junction rule is an application of the principle of conservation of electric charge: current is flow of charge per time, and if current is constant, that which flows into a point in a circuit must equal that which flows out of it.

What is an example of Kirchhoff's junction rule? ›

Kirchhoff's junction rule deals with how much current gets distributed when various branches of circuit meet. For instance, when you see a light bulb, you see that it lights up upon connecting to a battery.

What is a real life example of Kirchhoff's law? ›

Applications in daily life: In the deserts, days are very hot as sand is rough; therefore, it is a good heat absorber. Now by Kirchhoff's Laws, a Good absorber is a good emitter. So accordingly, the nights will be cool.

What can Kirchhoff's laws be used to solve? ›

Kirchhoff's laws are used for voltage and current calculations in electrical circuits.

### How is Kirchhoff's current law used? ›

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.

## Videos

1. Simplest Explanation of KIRCHHOFF'S LAWS (kcl kvl)
2. How to Solve a Kirchhoff's Rules Problem - Simple Example
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Author: Dong Thiel

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Introduction: My name is Dong Thiel, I am a brainy, happy, tasty, lively, splendid, talented, cooperative person who loves writing and wants to share my knowledge and understanding with you.