In 1845, Gustav Kirchhoff, a German physicist, developed a set of laws that explain conservation of energy and current within electrical circuits. This pair of laws then are collectively known as **Kirchhoff’s laws** or **Kirchhoff circuit law**.

The first **Kirchhoffs law** deals with voltage and is called Kirchoff’s voltage law. The second **Kirchhoffs law** deals with current and is called **Kirchhoff’s Current Law**. We will explain both in detail shortly. We will also include examples to make both laws more understandable.

## What Are **Kirchhoff’s Laws**?

At its core, the laws deal with the conservation of energy and current within electrical circuits. Since **Kirchhoff’s laws** deal with current and voltage, **electrical units of measure** used include Ampere and volt and Ohm. The followings are what each Kirchhoff law says about current and voltage:

- Kirchoff’s first law states that in a circuit, the total value of the currents in a node or junction is equal to the sum of the values of currents outside of the node or junction.
- Kirchoff’s second law states that the sum of the values of voltages around the closed-loop equals zero.

**Kirchhoff’s Current Law**

Ohm’s law can be applied in many cases. However, Ohm’s law does have its limitations. Although the law is excellent in the calculation of simple circuits (like **DC circuit**) whether they are series, parallel, or combinations of the two, it is inadequate in the calculation of complex circuits.

For complex circuits, another law is required. This law is **Kirchhoff’s Current Law**, also known as KVL, **Kirchhoff’s loop rule**, point rule, junction rule, or simply the First Law. This law can be used to calculate the currents or voltages of more complex circuits.

The first law of the **Kirchhoff’s laws** states that in a circuit, the total value of the currents in a junction is equal to the sum of the values of currents outside of the junction. To put it differently, the algebraic sum of the currents moving to or away from a node or junction is equal to zero or I(enter) + I(exit) = 0.

The value of the current can be either positive or negative. If the current is moving to a node or junction, the value is positive. If the current is moving away from a node or junction, the value is negative.

**Kirchhoff’s Current Law** Equation

Suppose that there are four currents in a node: I1, I2, I3, and I4. On one hand, I1, I2, and I3 enter the node and they are positive in value. On the other, I4 leaves the node and it is negative in value.

We can turn there into an equation, according to the first of **Kirchhoff’s laws**:

I1 + I2 + I3 – I4 = 0

It can also put like this:

∑I(entering) = ∑I(exiting)

Which means

∑I(Total) = 0

Note: a “node” in a circuit usually refers to a junction or connection of two or more current-carrying paths or elements. There has to be a closed-circuit path so the current can flow in and out of a node.

## Examples

#1 A simple junction with three currents

In a T-junction, 10A of current is moving into the node, and 4A of current is moving away from the node. 10A is I1 and 4A is I2. How much current is I3 and where does it flow?

Keep in mind that if a current is moving into the node, it has a positive value and if it moves away, it has a negative value.

Since ∑I(Total) = 0, that means

I1 + I2 + I3 = 0

I1 is moving into the node, which means it has a positive value

I2 is moving away from the node, which means it has a negative value

As such, the equation becomes

10 – 4 + I3 = 0

6 + I3 = 0

I3 = -6A

I3 has a negative value, which means it moves away from the node

This proves the first law of **Kirchhoff’s laws** which states that the total sum of the currents is 0. #2 A simple Junction with three currents

The junction is the same. I1 has a current of 10A and it moves into the node, I2 has a current of 14A and it moves away from the node. How much is the current of I3? In what direction does it move? Prove that the first law of **Kirchoff’s laws** that state the sum of the currents is equal to 0.

Keep in mind that if a current is moving into the node, it has a positive value and if it moves away, it has a negative value.

I1 + I2 + I3 = 0

10 – 14 + 13 = 0

-4 + I3 = 0

I3 = 4A

I3 has a positive value, which means it moves into the node

The total current moving into the node: 14A

The total current moving away from the node: -14A

This proves that ∑I(in) = ∑I(out) #3 A junction with four currents

In this junction, there are four currents: I1 with a current of 8A, I2 with a current of 9A, I3 with a current of 7A, and I4 current is unknown. I1 is moving away from the node while I2 and I3 are moving into the node. How much is the current of I4? To what direction does the current move?

Keep in mind that if a current is moving into the node, it has a positive value and if it moves away, it has a negative value.

I1 + I2 + I3 + I4 = 0

I1 moves away from the node, which means it has a negative value

I2 and I3 moves into the node, which means they have positive values

therefore,

-8 + 9 + 7 + I4 = 0

8 + I4 = 0

I4 = -8A

I4 has a negative value, which means it moves away from the node.

#4 A junction with five currents

Let’s try without a picture. Suppose a circuit has 5 branches. There are five currents: I1, I2, I3, I4, and I5. I1 has a current of 12A, I2 8A, I4 is twice I3 and I5 20A. I1 and I5 move into the node while I2 moves away. How much are the currents of I3 and I4? To what direction does it move?

Assume that if a current is moving into the node, it has a positive value and if it moves away, it has a negative value.

I1 + I2 + I3 + I4 + I5 = 0

12 – 8 + I3 + I4 + 20 = 0, where I4 is twice the amount of I3

12 – 8 + I3 + 2I3 + 20 = 0

4 + 3I3 + 20 = 0

24 + 3I3 = 0

3I3 = -24

I3 = -24 / 3

I3 = -8A

I4 = 2I3

I4 = 2 x -8

I4 = -16A

Both I3 and I4 have positive values, which means they move away from the node

**Kirchhoff’s Voltage Law**

The second law of **Kirchhoff’s laws** states that the sum of the values of voltages around the closed-loop equals zero. Put it differently, the algebraic sum (which means polarities, voltage drops, and signs of the sources are taken into account) of all the potential differences around any closed loop is equal to zero.

Kirchoff’s voltage law, just like the first law, is about the conservation of energy. Since the voltage moves around a closed circuit or loop, it will end up where it started. There is no voltage loss around the loop, which is why within the loop any voltage drops are equal to any voltage rises.

What is important to note is that the voltages inside a closed circuit can be either positive or negative. Determining which voltage is positive or negative is important. To determine whether a voltage is positive or negative, look at its current movement.

If the current is moving from high electric potential to low electric potential, the voltage is negative. The example for this is when current flows through a resistor, which reduces the energy. This results in a negative voltage.

Conversely, if the current is moving from low electric potential to high electric potential, the voltage is positive. For example, when current moves from low potential to high potential, which means the energy is increased. This results in a positive voltage.

There many ways to make use of **Kirchhoff’s Voltage Law**. For example, both loop analysis and **mesh analysis **make use of **Kirchhoff’s Voltage Law** if a given circuit has a solution. In both cases, the first **Kirchhoffs law** is used to arrive at a solvable set of equations.

**Kirchhoff’s Voltage Law** Equation

Suppose a closed circuit has 4 points: A, B, C, and D. Between two points there is a resistor so, in total, there are 4 resistors in the circuit. The equation according to **Kirchhoff’s Voltage Law** is:

VAB + VBC + VCD + VDA = 0

## Examples

#1: A circuit with three resistors

A circuit with a voltage of 12V has three resistors: R1, R2, and R3. The resistance of R1 is 8Ω. R2 10Ω., and R3 12Ω. The electrical charges move from low potential of the battery to high potential, which means VDA is +12V. Prove the second of **Kirchhoff’s laws**!

VDA + VAB + VBC + VCD = 0

where

- VAB is the voltage drop from point A to point B
- VBC is the voltage drop from point B to point C
- VCD is the voltage drop from point C to point D
- VDA is the voltage rise from point D to point A

Voltage drop has negative value while the voltage rise has positive value so the formula becomes

+VDA – VAB – VBC – VCD = 0

To find the voltage drops, we can use I = V / R

12V – (I x 8) – (I x 10) – (I x 12) = 0

12V – 8I – 10I – 12I = 0

12V – 30I = 0

12V = 30I

I = 12V / 30

I = 0.4A

Thus

12 – (0.4 x 8) – (0.4 x 10) – (0.4 -12) = 0

12 – 3.2 – 4 – 4.8 = 0

This proves the Kirchoff’s Voltage Law as the sums of the voltages equal to zero

#2: A circuit with two batteries and two resistors

This circuit has two currents and two resistances. The currents are I1, which has 12V of current, and I2 has 8V of current. The resistances are R1, which has 50Ω of resistance, and R2 has 30Ω of resistance. The current moves in a clockwise direction. How much is the potential at every point?

Using the Kirchoff’s Voltage Law,

VB1 + V1 + VB2 – V2 = 0

where

- VB1 is the voltage rise from the first battery
- VB2 is the voltage rise from the second battery
- V1 is the voltage drop from point A to point B
- V2 is the voltage drop from point C to point D

since the current from the batteries move from lower potential to higher potential, they have positive values while the current that moves through a resistor has a negative value

Therefore the formula becomes

+VB1 – V1 + VB2 – V2 = 0

to find V1 and V2, use V = I x R

so

12 – (I x 50) + 8 – (I x 30) = 0

12 – 50I + 8 – 30I = 0

20 – 80I = 0

-80I = -20

I = – 20 / – 80

I = 0.25A

Putting the current into the **Kirchhoff’s Voltage Law** equation

12 – (0.25 x 50) + 8 – (0.25 x 30) = 0

12 – 12.5 + 8 – 7.5 = 0

20 – 20 = 0

Now, to find the potential at each point. The circuit has 4 points: A, B, C, D

The potential at point B is 12V due to the first battery

The potential difference in point C is

V1 = VB1 – I x R1

V1 = 12 – (0.25 x 50)

V1 = 12 – 12.5

V1 = 0.5V

The potential at point C is 0.5V

The potential at point D is 8V – 0.5 = 7.5V

The potential at point A is

V2 = VB2 – I x R2

V2 = 7.5 – 0.25 x 30

V2 = 7.5 – 7.5

V2 = 0V This proves the second of **Kirchhoff’s laws** that the sum of all the voltages is zero.

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