Kirchoff’s Laws 1 and 2: Sounds, Formulas & Applications in Everyday Life

Kirchoff’s Laws – In Physics, Sinaumed’s definitely learns about electricity or electricity. In this chapter, you will usually get acquainted with Kirchoff’s Laws 1 and 2, which discuss the matter of electrical conduction. Kirchoff’s law is used as a basis for understanding electric circuits, especially closed electricity.

Yup , by understanding Kirchoff’s laws Sinaumed’s can analyze various simple electrical circuits by changing the series and parallel combinations of resistors present in a circuit with many resistors in it.

Then, what exactly is meant by Kirchoff’s laws 1 and 2? Let’s discuss together here.

Acquaintance with the Inventor of Kirchoff’s Law: Gustav Robert Kirchhoff

Gustav Robert Kirchoff was a German physicist born on March 12, 1824 who died at the age of 63, on October 17, 1887 to be precise. Kirchoff was born to Friedrich Kirchoff and Johanna Henriette Wittke. Friedrich, his father, was a lawyer by profession at that time.

At the age of 23, Kirchoff married a woman named Clara Richelot. From this marriage, he has two daughters and three sons. In 1869, Richelot died and Kirchoff finally married again to Luise Brommel in 1872.

Part of Kirchoff’s life was spent as a professor and professor of physics. From this dedication, he succeeded in discovering circuit law formulas for electrical engineering, the laws of thermodynamics or thermal radiation, and contributed to the field of spectroscopy.

You can find out a more complete biography of Gustav Robert Kirchoff and other physicists in the book Encyclopedia of 20th Century Physicists written by Taufik Hidayat.

Kirchhoff’s Law 1

Kirchoff’s law 1 is also known as the junction rule or the law of branching which fulfills the conservation of charge. This law is usually used in multi-symptom circuits and has several branching points that divide the electric current.

In a constant state, there is no accumulation of electric charge at any point in the circuit. This means, the amount of charge that goes into each point is equal to the amount of charge that comes out of that point.

This is in line with the sound of Kirchoff’s law 1 which says:

“The sum of the electric currents going into a branch point will equal the sum of the currents leaving that point.”

To understand better, take a look at the image below.

The basis for the sound of this law is taken from the Law of the Conservation of Charge which has the sound “The charge at point A will be the same as the charge at Point B”. But with notes, there are no external forces acting in this series.

It’s like a tourist bus carrying 50 people from Bandung to Jakarta. Since departing from Bandung to arriving at their destination in Jakarta, the number of passengers on the bus remained 50. With a note, no passengers were left behind during breaks at rest areas and other incidents.

Kirchoff’s Law Formula 1

Kirchoff’s law formula is as follows:


Σ Input = Total incoming current (Ampere)
Σ Iout = Total current going out (Ampere)

So, so that Sinaumed’s can understand more about this Kirchoff 1 law formula, let’s practice by understanding the example questions below.

Look at this picture of the electrical circuit:

What amount of current is there in I4?

Total incoming current = total outgoing current. So:
I1 + I4 = I2 + I3
5 A + I4 = 10 A + 2 A
I4 = 12 A – 5A
I4 = 7 A
So, the answer to the question is 2 A.

Kirchhoff’s Law 2

Kirchoff’s law 2 or the loop rule has a potential difference between two branching points in a circuit that is in a constant state.

This law is proof of the law of conservation of energy. Where if the charge Q that exists at any point has a potential V, then the energy possessed by that charge is QV.

Then, as the charge begins to move through the loop, it gains or loses energy as it passes through an element. However, when the charge returns to its starting point, the energy in the charge returns to QV.

Here is an explanation of Kirchoff’s law 2 which reads:

“The sum of the total voltages around the closed loop in a network, has a magnitude equal to zero”.

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Usually Kirchoff’s laws discuss loops in an electric circuit like the one in the picture below. The goal is to measure how much the voltage potential difference is in a circuit that has no branches.

Kirchoff’s Law Formula 2

Kirchoff’s 2nd law formula is as follows:

Kirchoff’s Rule of Law

Apart from the formula, you also have to pay attention to some rules from Kirchoff’s law which will help you understand it. These rules are as follows:


For each closed path, Sinaumed’s must choose a loop with a certain direction. You are free to determine the direction of the loop (whether you want it clockwise or counterclockwise), but it’s better to use the direction that goes with the current.


If the direction of the loop in a branch is the same as the direction of the current, then the voltage drop (IR) will have a positive sign. If the direction is opposite, then the sign becomes negative.

For example, the direction of the loop and the current is clockwise, so the voltage drop is positive.

Loop in the direction of current


If when following the direction of the loop, the pole of the voltage source encountered for the first time is the positive pole, then the Electric Force Movement (EMF) will have a positive sign.

Application of Kirchoff’s Laws in Everyday Life

Generally, to understand the laws that exist in Physics, we are invited to see their application in everyday life. This method will be easier to understand because we experience it ourselves and understand every change and how it works.

It’s different if you only learn from textbooks. Sometimes the sound of the law is difficult to understand. Especially when added with certain formulas. The more difficult it is for us to understand.

Unfortunately, when studying Kirchoff’s laws you cannot rely on this strategy. This is because the application of Kirchoff’s laws in everyday life is quite difficult for our eyes to see directly.

Its application is in the form of a formula used to run electrical circuits, namely series circuits and parallel circuits. The easiest example to describe these two circuits is the lights in the house.

If you use a series circuit at home, the lamp that is closest to the switch or power source will definitely glow brighter than the lamp that is far from the power source.

If what is used for the lights in your house is a parallel circuit, then all the lights will light up with the same level of brightness regardless of the difference in distance between the lights and the power source.

So the question is, when do we actually have to analyze a circuit? Then who can analyze it? You can see the answer below:

  1. We analyze circuits when we want to check for damage to a device.
  2. We analyze circuits when we build our own electronic circuits by studying pre-existing circuits.

Kirchoff’s law itself has advantages, which can be used to analyze very complicated circuits. For example, when Sinaumed’s wants to replicate an electronic circuit that has a different specification, you have to replace some of the components in it. Either that, increase or decrease the resistance value of the resistor.

So that you can determine the appropriate component, you must calculate the value of the voltage and current in the circuit.

Example of Kirchoff’s Law Problem 1 and Discussion

Problem 1

In the simple circuit below, the current flowing in I1 = 20 A, I2 = 4 A, and I4 = 8 A. Determine the value of I3!

It is known:
I1 = 20 Amperes
I2 = 4 Amperes
I4 = 8 Amperes

Asked: I3?


According to the data that has been mentioned, this problem can be solved using the Kirchoff 1 law formula, namely:

ΣImasuk = ΣIkeluar
I1 = I2 + I3 + I4
20 = 4 + I3 + 8
20 = 12 +I3
I3 = 20-12= 8 Ampere

So the answer to the question is 8 Amperes.

Problem 2

In the simple circuit below, the current flowing at I1 = 15 Amperes, I3 = 7 Amperes, I4 = 8 Amperes and I5 = 5 Amperes. Determine the value of I2!

I1 = 15 Ampere
I3 = 7 Ampere
I4 = 8 Ampere
I5 ​​= 5 Ampere

Asked: I2?


According to the data that has been mentioned, this problem can be solved using the Kirchoff 1 law formula, namely:

ΣImasuk = ΣIkeluar
I1 + I2 = I3 + I4 + I5
20 + I2 = 7 + 8 + 5
I2 = 20-15
I2 = 20-15= 5 Ampere

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So the strong current on I2 is 5 Amperes.

Problem 3

If I1 in question number one is increased by 25%, determine the ratio of the initial and final I3 values.


I1 = 20 Ampere
I2 = 4 Ampere
I4 = 8 Ampere
I3 = 8 Ampere
I1’= 20 + (20 x 0.25) = 25 Ampere

Ditanyakan: I3:I3′?


According to existing data, this problem can be solved using Kirchoff’s 1st law formula:

ΣImasuk = ΣIkeluar
I1′ = I2 + I3′ + I4
25 = 4 + I3′ + 8
25 = 12 +I3′
I3′ = 25-12= 13 Ampere
I3:I3’= 8:13

So the initial and final I3 comparison is 8:13.

Example of Kirchoff’s Law Questions 2 and Discussion

Problem 1


R1 = 3 ohms
R2 = 2 ohms
R3 = 1 ohm
є1 = 12 Volts
є2 = 24 Volts

Asked: I?


According to existing data, this problem can be solved using the Kirchoff 2 law formula:


First, determine the direction of the loop first. To make it easier to do, the direction of the loop is the same as the direction of the electric current (I)

To make it easier to calculate and write, they are written sequentially

Σє + ΣIR = 0
IR2 + є1 + IR1 + IR3 -є2 = 0
I(2) + 12 + I(3) + I(1) -24 =0
I=12/6= 2 Ampere

So the current strength in I is 2 Amperes.

Problem 2

If it is known that ε1 = 16 V; ε2 = 8V; ε3 = 10V; R1 = 12 ohms; R2 = 6 ohms; and R3 = 6 ohms. How big is the electric current I?

ε1 = 16 V
ε2 = 8 V
ε3 = 10 V
R1 = 12 ohm
R2 = 6 ohm
R3 = 6 ohm.

Asked: I?


First, determine the direction of the loop first.

The calculation starts from R1

Σє + ΣIR = 0
IxR1 – ε1 + I1xR2 = 0
I(6) – 16 + I1(6) = 0
12I + 6I1 = 16 ….. > dibagi 2
6I + 3I1 =8
karena I1+I2=I, maka
6(I1+I2) +3I1 = 8
6I1 + 6I2 + 3I1 = 8
9I1 + 6I2 = 8

Loop 2 which is on the right
In loop 2, the direction of current I1 is opposite to the direction of the loop until, I1 is negative (-), until

Σє + ΣIR = 0
I2xR3 + ε2 – I1xR2 + ε3 = 0
I2(6) + 8 – I1(6) + 10 = 0
6I2 – 6I1 + 18 = 0
– 6I1 + 6I2 = -18


9I1 + 6I2 = 8
– 6I1 + 6I2 = -18
—————– (-)
15I1 + 0 = 26
I1 = 26/15 A
-6I1 + 6I2 = -18 …..> -6(26/15) + 6I2 = -18…> I2=12/5 A
I=I1+I2= (26/15) + (12/5) = 4,13 A

so the current strength in I is 4.13 Amperes.

Problem 3

A circuit is known as below. What will happen to the electric current in the circuit if switch S is closed. Then what is the power at R = 2 Ω? when R1 = 2 Ω, R2 = 4 Ω, R3 = 1 Ω, є1 = 12 V, є2 = 8 V


First determine the direction of the current and the direction of the loop. In this case the direction of the current is the same as the direction of the loop so that The image above becomes

Loop 1

ε + IR + 0 or Φ +ΙR = 0
-ε1 + I1 R2 + I R1 =0
-12 + 4I1 + 2I=0
2I1 +2I2+4I1=12
6I1+2I2=12 :2 = 3I1+I2=6

Loop 2

8 – I1 R2 +I2 R3=0
8 – I1 4 + I2 1=0


———- (+)
I2=6-6=0 Ampere
I=I1+I2=2+0=2 Ampere

Power on R1

=(2^2)2=8 Watt

So the power on R2 is 8 Watts

Problem 4

Determine the value of the current strength of the circuit below and the clamping voltage at the KJ point. If it is known, ε1= 5 volts, r1 = 1 Ω, R1= 2 Ω, ε2= 17 volts, r2 = 1 Ω, R2= 2 Ω.


Wanted: Determine the value of the current strength (I) of the circuit below and the clamping point voltage KJ(V)


The first step is to determine the direction of the current and the direction of the loop. To make it easy to direct current and unidirectional loops. Like the picture below.

VKJ= ε2+Ir2
VKJ= 17+1=18 Volts

The current strength (I) of the circuit is 1 Ampere and the KJ(V) point clamp voltage is 18 volts

All of the questions above were inspired by examples of questions in the Class XI SMA/MA Fundamental Physics book by Sunardi. This book presents regular physics questions specifically designed to be used as a practice ground for students in preparing themselves for various forms of physics exams, such as exams at the school level, national assessments, to state university entrance selection (PTN).

Thus the discussion of Kirchoff’s laws 1 and 2 along with examples of problems. After reading this article to the end, I hope it can add insight and be useful for Sinaumed’s.