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. If you dive deeper, you will see that wires connect the bulb and battery plus a number of tiny electrons moving around.

Similarly, the battery is responsible for giving electrons energy to put through the filament of the bulb. When moving through the filament, they slow down and some energy of theirs transforms into light. Thus, you might wonder are electrons disappearing in the bulb or are fewer coming out than going inside.

Well, the answer is the same number is coming out. The change is only in the energy of theirs. Thus, when electrons charge, the total amount of charge conserves. Therefore, as the charge always conserves, we can determine how current should flow in each branch of a circuit. This article deals with the junction rule and its application.

**Definition**

A circuit consists of places known as junctions. Over here, various wires come together. As the charge conserves and current measures the rate at which these charges flow, the total current that comes out in the junction equals the total current which comes out the other side of the junction. This is similar to the light bulb we discussed above. Therefore, this relationship can be called Kirchhoff’s Junction Rule.

In other words, we see that Kirchhoff’s junction rule is applying the principle of preservation of electric charge: current is the flow of charge per time, and then if the current is continuous, that which flows into a point in a circuit must equal to that which flows out of it.

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## Kirchhoff’s Junction Rule

A closes circuit can have several circuit elements like batteries and resistors. The circuit branches producing junctions, where the circuit either separates or recombines. The sum of the current in and out of the circuit junction should be 0. We measure current in Amperes (A).

\(\sum I\) = 0

Over here:

\(\sum I\) is the sum of currents in or out of a circuit junction which is 0

I refer to the current (Amperes – A)

**Solved Example**

**Question- **There is a circuit which comprises two resistors plus a voltage source (battery). Thus, the current before junction “a” is I_{a}, and the current through the resistor R_{1} is I_{1}. Furthermore, the current through resistor R_{2} is I_{2. }You need to find out the value of current I_{1 }when the value for I_{a} is 5.50 A and I_{2 }is 2.00 A.

**Answer- **Following Kirchhoff’s Junction Rule, we see that the sum of currents in and out of the junction should be 0. Therefore, in this scenario, I_{3} connects to junction “b”. Further, the current flows through circuit in a clockwise direction, so remember these directions of current are vital. We can use the sum of the currents flowing in and out of junctions “a” and “b” to find out the value of I_{3}. Therefore, the sum of current coming in and out of junction “a” is:

\(\sum I\) = 0

\(\sum I\) = \(I_{a} – I_{1} – I_{2}\) = 0

(3.50 A) – \(I_{a}\) = 0

(3.50 A) – I_{1 }= 0

Therefore, in order to find the value of \(I_{1}\), we will rearrange the above formula:

(3.50 A) – \(I_{1}\) = 0

– \(I_{1}\) = – (3.50 A)

\(I_{1}\) _{ }= 3.50 A

Thus, we see that the value of current \(I_{1}\) 3.50 A(Amperes).

Typo Error>

Speed of Light, C = 299,792,458 m/s in vacuum

So U s/b C = 3 x 10^8 m/s

Not that C = 3 x 108 m/s

to imply C = 324 m/s

A bullet is faster than 324m/s

I have realy intrested to to this topic

m=f/a correct this

M=f/g

Interesting studies

It is already correct f= ma by second newton formula…