Kirchhoff’s Law, Ohm’s Law, along with the questions and discussion – Some of you have probably heard or read the name Kirchhoff , right? Yep , Kirchhoff is one of the physicists who has contributed to the science of electricity. The laws of his famous research results are Kirchhoff’s Laws I and II.
Kirchhoff’s law was coined by Gustav Robert Kirchhoff who was a physicist from Germany. Kirchhoff explained his laws about electricity into two parts, namely Kirchhoff’s First Law and Kirchhoff’s Second Law.
Meanwhile, Ohm’s law is a law regarding electricity that was coined by a physicist who came from Germany as well, namely Georg Simon Ohm.
Then, what is the relationship between these 2 laws?
KIRCHOFF’S LAW
As the name implies, the sound of Kirchoff’s Laws was coined by a physicist from Germany named Gustav Robert Kirchoff. Gustav Kirchoff’s research discusses the subject of electrical conduction.
Quoting from Merriam-Webster, the definition of Kirchoff’s current law is:
In an electric circuit, the sum of the currents that exist in all branches and meet at any point is zero.
Besides that, Merriam-Webster also defines Kirchoff’s Law as follows:
In a closed electric circuit, the sum of the electromotive forces in the circuit is equal to the resistance times the current.
Then, what is the sound of Kirchoff’s Laws 1 and 2? Check out the explanation below:
Kirchoff’s Laws 1
In everyday life, we often encounter electrical circuits consisting of various connections. This means that the electric circuit has many branches and nodes.
This node is the result of a meeting of three or more branches. Pay attention to the following picture:
The sum of the currents entering the node is equal to the sum of the currents leaving. This law is also known as the law of conservation of electric charge.
Quoting from Acceleration Physics, the sound of Kirchoff’s 1st Law is:
“The sum of the electric currents entering a node is equal to the sum of the electric currents coming out of the node.”
Mathematically, Kirchoff’s Law 1 can be written with the formula:
So that You can understand more about Kirchoff’s Law 1, let’s practice the example questions below.
Look at this picture of the electrical circuit:
The current flowing in I1 = 5 A, I2 = 10 A, and I3 = 2 A, what is the total current in I4?
Completion:
If the sum of the incoming currents equals the outgoing currents, then:
I1 + I4 = I2 + I3
5 A + I4 = 10 A + 2 A
I4 = 12A – 5A
I4 = 7 A
So, the total strong current on I4 is 2A.
Kirchoff’s 1st law can be combined with series and parallel electrical connections in Ohm’s Law.
The series connection has the purpose of increasing the resistance in the circuit and as a voltage divider. In a series connection, the current through each resistor has the same value.
Meanwhile, for parallel electrical connections, the potential difference (V) or the voltage through each resistor is the same.
Parallel electrical connections have the purpose of minimizing the resistance that exists in the circuit and as a current divider.
Kirchoff’s Law Sounds 2
Kirchoff’s Law 2 has a function for more complicated electric circuits, which cannot be simplified using Kirchoff’s Law 1. This electric circuit is usually a closed electrical circuit or what is commonly called a loop.
In a closed electrical circuit, it does not only use one source of voltage or electromotive force (EMF), but can have two or more.
In a closed circuit, the algebraic sum of the electromotive forces with a voltage drop equals zero.
Mathematically, the formula for Kirchoff II’s Law can be written as follows:
When solving a closed electrical circuit using Kirchoff’s 2nd Law, use the following steps:
Selects the loop direction at each of the closed paths. The choice of the direction of the loop is free, but usually the direction of the loop is in the direction of the current in order to make it easier to work on.
The voltage drop (IR) has a negative value if the loop direction is opposite to the current direction. The voltage drop (IR) will have a positive value if the loop direction is in the same direction as the current.
When following the direction of the loop and the voltage source encountered first is the positive pole, the EMF also has a positive sign, and vice versa.
OHM’S LAW
In Electronics, the basic Law of Electronics that every Electronics Engineer or hobbyist must learn and understand is Ohm’s Law, which is the basic law that states the relationship between Electric Current (I), Voltage (V) and Resistance (R). Ohm’s Law in English is called “Ohm’s Laws”. Ohm’s Law was first introduced by a German physicist named Georg Simon Ohm (1789-1854) in 1825. Georg Simon Ohm published Ohm’s Law in a Paper entitled “The Galvanic Circuit Investigated Mathematically” in 1827.
Ohm’s Law sounds
Basically, the sound of Ohm’s Law is:
“The amount of electric current (I) flowing through a conductor or conductor will be directly proportional to the potential difference / voltage (V) applied to it and will be inversely proportional to the resistance (R)”.
Mathematically, Ohm’s Law can be formulated into the following equation:
V = I x R
I = V / R
R=V/I
Where :
V = Voltage (Potential Difference or Voltage whose unit is Volt (V))
I = Current (Electric Current whose unit is Ampere (A))
R = Resistance (Barriers or Resistance whose unit is Ohm (Ω))
In its application, we can use Ohm’s Law Theory which is in Electronic Circuits in order to minimize electric current, voltage, and also to obtain the appropriate resistance (resistance value).
The thing to remember in calculating the Ohm’s Law formula, the units used are Volts, Amperes and Ohms. If we use other units such as millivolts, kilovolts, milliamperes, megaohms or kiloohms, then we need to convert to Volts, Amperes and Ohms units first to simplify calculations and also to get the correct results.
Case Examples in Ohm’s Law
To make it clearer in understanding Ohm’s Law, we can do practical work using a simple electronic circuit like the following:
This practicum requires a DC Generator (Power Supply), Voltmeter, Ammeter, and a Potentiometer according to the required value.
From a picture of a simple electronic circuit, we can compare Ohm’s Law Theory with the results obtained from the Practicum in terms of calculating Electric Current (I), Voltage (V) and also Resistance / resistance (R).
Calculating Electric Current (I)
The formula that can be used to calculate Electric Current is I = V / R
Example Case 1:
Setting the Power Supply or DC Generator used to produce a 10V Output Voltage, then set the Potentiometer Value to 10 Ohms. What is the value of Electric Current (I) ?
Enter the Voltage value, which is 10V and also the Resistance Value of the Potentiometer, which is 10 Ohm, into the Ohm’s Law Formula as follows:
I = V / R
I = 10 / 10
I = 1 Ampere
So the result is 1 Ampere.
Example Case 2:
Setting the DC Generator or Power Supply to produce a 10V Voltage Output, then set the Potentiometer value to 1 kiloOhm. What is the value of Electric Current (I)?
First convert the resistance value of 1 kiloOhm to Ohm units. 1 kiloOhm = 1000 Ohms. Then, input the 10V Voltage value and also the Resistance value from the 1000 Ohm Potentiometer into the Ohm’s Law Formula as follows:
I = V / R
I = 10 / 1000
I = 0.01 Amperes or 10 milliAmperes
Then the result is 10mA
Calculating Voltage (V)
The formula that will be used to calculate Voltage or Potential Difference is V = I x R.
Sample case :
First, set the value of the resistance or resistance (R) Potentiometer to 500 Ohm, then set the DC Generator (Power supply) to get the amount of Electric Current (I) 10mA. What is the Voltage (V)? First convert the unit of Electric Current (I), which is still one milliAmpere, into Ampere units, namely: 10mA = 0.01 Amperes. Then, input the Potentiometer Resistance value of 500 Ohm and the Electric Current value of 0.01 Amperes into the Ohm’s Law Formula as follows:
V = I x R
V = 0.01 x 500
V = 5 Volts
Then the value is 5Volt.
Calculating Resistance / Resistance (R)
The formula that will be used to calculate the Resistance Value is R = V / I
Sample case :
If the Voltage value on the Voltmeter (V) is 12V and the Electric Current value (I) on the Ammeter is 0.5A. So, what is the value of the resistance on the potentiometer?
First, input the value of 12V Voltage and 0.5A Electric Current into the Ohm Formula as follows:
R = V / IR
= 12 /0.5
R = 24 Ohm
Then the resistance value is 24 Ohm
Questions and Discussion
1. In the simple circuit example in the previous figure, the current flowing at I1 = 20 Amperes, I2 = 4 Amperes and I4 = 8 Amperes.
Determine the value of I3.
Discussion:
It is known that:
I1 = 20 Ampere
I2 = 4 Ampere
I4 = 8 Ampere
Asked: I3 =… ?
Answer :
Based on the data contained in question no. 1 can be solved by Kirchoff’s law 1 namely,
ΣI in = ΣI out
I1 = I2 + I3 + I4
20 = 4 + I3 + 8
20 = 12 +I3
I3 = 20-12= 8 Amperes
Then it is obtained that the current strength on I3 is 8 Amperes.
2. In this simple circuit, the current flowing at I1 = 15 Amperes, I3 = 7 Amperes, I4 = 8 Amperes and I5 = 5 Amperes. Determine the value of I2.
Discussion:
Is known :
I1 = 15 Amperes
I3 = 7 Amperes
I4 = 8 Amperes
I5 = 5 Amperes
Asked: I2 =… ?
Answer :
Based on the existing data in question no. 2 can be solved by Kirchoff’s law 1 namely,
ΣIn = ΣI out
I1 + I2 = I3 + I4 + I5
20 + I2 = 7 + 8 + 5
I2 = 20-15
I2 = 20-15 = 5 Amperes
Then it is obtained that the large current strength on I2 is 5 Amperes.
How much current flows in an electric current circuit if R1 = 3 ohms, R2 = 2 ohms, and R3 = 1 ohm and є1 = 12 Volts and є2 = 24 Volts.
Discussion:
Is known :
R1 = 3 ohms
R2 = 2 ohms
R3 = 1 ohm
є1 = 12 Volts
є2 = 24 Volts
Asked: I =… ?
Answer :
Based on the existing data, question no. 2 can be solved using Kirchoff’s 2nd law
Step:
First, determine the direction of the loop. To make it easier, the direction of the loop is the same as the direction of the electric current (I)
From the picture, the electric current (I) meets the pole (+) at є1 so that it has a positive value є1 (+) while є2 has a negative value є2 (-)
In order to make it easier to calculate and also write, it is written sequentially Σє + ΣIR = 0
IR2 + є1 + IR1 + IR3 -є2 = 0
I(2) + 12 + I(3) + I(1) -24 =0
6I-12=0
I=12/6= 2 Amperes
Then it is obtained that the large current strength at I is 2 Amperes.
- If it is known that ε1 = 16 V; ε2 = 8V; ε3 = 10V; R1 = 12 ohms; R2 = 6 ohms; and R3 = 6 ohms. So, the magnitude of the electric current I is…
Discussion:
Is known :
ε1 = 16 V
ε2 = 8 V
ε3 = 10 V
R1 = 12 ohms
R2 = 6 ohms
R3 = 6 ohms.
Asked: I =… ?
Answer:
The calculation starts from R1
Σє + ΣIR = 0
IxR1 – ε1 + I1xR2 = 0
I(6) – 16 + I1(6) = 0
12I + 6I1 = 16 …..> divided by 2
6I + 3I1 = 8
because I1+I2=I
so
6(I1+I2) +3I1 = 8
6I1 + 6I2 + 3I1 = 8
9I1 + 6I2 = 8
Loops 2
In loop 2, the direction of current I1 is opposite to the direction of the loop so that, I1 is negative (-), so
Σє + ΣIR = 0
I2xR3 + ε2 – I1xR2 + ε3 = 0
I2(6) + 8 – I1(6) + 10 = 0
6I2 – 6I1 + 18 = 0
– 6I1 + 6I2 = -18
Elimination
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
Then it is obtained that the current strength at I is 4.13 Amperes.
That’s all, a brief summary of Kirchoff’s law and Ohm’s law, thank you for reading and I hope this is useful for all of you.