# Current Divider

 DC Devre Analizi DC Devre Analizi Ohm Kanunu ve Güç Elektrik Ölçü Birimleri Kirşof Devre Kanunları Mesh(Çevre Akımları) Analizi Node(Düğüm Gerilim) Analizi Thevenin Teoremi Norton Teoremi Maksimum Güç Transferi Yıldız Delta Dönüşümü Voltaj Kaynakları Akım Kaynakları Kirchhoff'un Gerilim Kanunu Kirchhoff'un Akım Kanunu Gerilim Bölücüler Akım Bölücüler Elektrik Enerjisi ve Güç

In current divider circuits, there are two or more parallel arms for the flow of currents, but the voltage is the same for all components in the parallel circuit.

Current Dividing Circuits areparallel circuits in which the welding or supply current is divided into several parallel paths.In a parallel connected circuit, the terminals of all components that share the same two end nodes are connected to each other.This results in different paths and branches for the current to flow or pass through.However, currents can have different values over each component.

The main feature of parallel circuits is that the voltage is common to all connected roads, although they can produce different currents flowing from different branches.So V R1 = V R2 = V R3 … etc. Therefore, kirchhoff's current law (KCL) and of course the Ohm Act eliminate the need to find individual resistance voltages by allowing the easy availability of branch currents.

## Resistant Voltage Divider

The easiest and most basic form of passive current divider network is two resistances connected in parallel.The Current Divider Rule allows us to calculate the current flowing from each parallel resistance as a percentage of the total current.Consider the circuit below.

### Resistant Current Dividing Circuit

Here, this basic current dividing circuit consists of two resistances: R1 and R2, in parallel, divide the supply or welding current into two separate currents between them, IR1 and IR2,and come together again and return to the source.

Since the welding or total current is equal to the sum of individual branch currents, the total current flowing in the circuit is given by the Kirsof current rule KCL as follows:

I T = I R1 + I R2

Since the two resistors are connected in parallel, Kirchhoff's Currents Act (KCL) must follow that the current flowing from R1will be equal to:

I R1 = I T – I R2

and the current flowing from resistance R2will be equal to:

I R2 = I T – I R1

Since the same voltage (V) is present throughout each resistant element, following the Ohm law we can find the current flowing from each resistance in terms of this common voltage, since it is simply V = I*R. Therefore, for voltage (V) along the parallel combination:

Solving for IR1 provides:

Similarly, solving for IR2 gives:

Note that there is contrasting resistance in the share of the above equations for each branch current.So we use R2 to solve I1and R 1 to solve I 2. This is due to the fact that each branch is inversely proportional to the resistance of the current, which results in smaller resistance with a larger current.

### Current Divisive Question Example 1

20Ω resistance is connected in parallel with 60Ω resistance.If the combination is connected to a 30 volt power supply, find the current that passes through each resistance and the total current provided by the source.

Note that the smaller 20Ω resistance has a larger current, since by its nature, a larger current will always flow from the least resistant path or branch.This means that the short circuit will produce maximum current flow, while the open circuit will result in zero current flow.Also note that the equivalent resistance of parallel-linked resistors, R EQwill always be less than the omic value of the smallest resistance, and the equivalent resistance decreases as more parallel resistance is added.

Sometimes it is not necessary to calculate all branch currents, if the feed or total current, I T is already known, then the final branch current can be found by removing the calculated currents from the total current defined by the Kirchhoffs current law.

### Current Divisive Question Example 2

As shown below, three resistors are connected to form a current dividing circuit.If the circuit is powered by a 100 volt 1.5kW power supply, calculate independent branch currents using the current splitting rule and equivalent circuit resistance.

1) Total circuit current I T

2) Equivalent resistance R EQ

3) Branch currents I R1 , I R2 , I R3

We can control our calculations according to Kirchhoff's current rule, that is, all branch currents will be equal to the total current, that is: IT = IR1 + IR2 + IR3 = 10 + 4 + 1 = 15 amps as expected. Thus, we can see that the total current is divided according to a simple ratio determined by branch resistors. In addition, as the number of resistances connected in parallel increases, the feeding of the total current will also increase for a certain feeding voltage.

## Current Division Using Conductors

Another simple method of finding branch currents in a parallel circuit is to use the method of conductivity.In DC circuits, conductivity is the equivalent of resistance and is indicated by the letter " G ".Since conductivity (G) is the equivalent of resistance (R) measured in Ohm (Ω), ohm's equivalent is called "mho" (℧), (an inverse ohm sign).So G = 1/R .The electrical units given to the conductivity are Siemen (symbol S).

Therefore, equivalent or total conductivity for parallel connected resistors, CT, will be equal to the sum of individual conductivity as shown.

### Parallel Conductivity

Therefore, if the constant value of a resistance is 10Ω, its equivalent conductivity will be 0.1S, etc.Since it is reciprocal, a high conductivity value represents a low resistance value, and vice versa.We can also use milli-Siemens, mS, micro-Siemens, uS and even nano-Siemens , nS prefixes for very small conductivity.So a resistance of 10kΩ will have 100uS conductivity.

Using the Ohm Law equation for current with I = V/R, we can define branch currents using conductivity as follows: I = V*G

In fact, we can take this a step further by saying that the feed current to our parallel resistant network above is as follows:

But we know from above that the voltage for a parallel connected circuit is common to all components, and since the voltage is equal to the current multiplied by resistance V = I*R, therefore, we can conclude that the voltage is equal to the per-segment conductivity of the current when using conductivity. .This is V = I/G.

Next, we can state the above equations for the current divisive rule regarding conductivity (G) instead of resistance (R):

### Current Divider Rule Using Conductivity

Similarly, parallel resistors are given as R2 and R3 for currents:

Unlike the above equations for resistance, you may have noticed that each branch has the same conductivity in its share of the current.So we use G 1to solve I 1and G 2to solve I 2.This is due to the fact that the conductivity is mutual resistance.

### Current Divisive Question Example 3

Using the conductivity method, locate the individual branch currents of the following parallel resistance circuit, I 1 , I 2, and I 3.

Total conductivity G T

Total feed current I S

Individual branch currents I 1 , I 2 and I 3

Since conductivity is mutual or vice versa of resistance, the equivalent resistance value of the sample circuit is simply 1 / 800us, which is equal to 1250Ω or 1.25 kω, which is clearly less than the smallest resistance value of R1 at 2kω.

## Current Divider Summary

Current dividers arethe process of finding individual branch currents in a parallel circuit where each parallel element has the same voltage. Kirchhoff's current law (KCL) states that the algebraic sum of individual currents entering a port or node will be equal to the currents that leave it.So the net result is zero.

Kirchhoff's current divider rule can also be used to find individual branch currents when equivalent resistance and total circuit current are known.In the case of only two resistant branches, the current in a branch will be part of the total current I T.If the two parallel resistant branches are equal in value, the current will be divided evenly.

In the case of three or more parallel branches, equivalent resistance R EQ is used to divide the total current into fractional currents for each branch, which produces an equal current ratio, which, contrary to the resistance values, has a smaller resistance value.Supply or total current, I T is the sum of all individual branch currents.This makes current dividers useful for use with existing resources.

It is sometimes convenient to use conductivity with parallel circuits, as it can help reduce the mathematics required to determine branch currents through separate circuit elements that are connected to each other in parallel.This is because the total conductivity for parallel circuits is the sum of individual conductivity values.Conductivity is the opposite or opposite of resistance in G = 1/R.The conductivity units are Siemens, S. The conductivity of an element can be used for current dividers even if the supply voltage is DC or AC.

 DC Devre Analizi DC Devre Analizi Ohm Kanunu ve Güç Elektrik Ölçü Birimleri Kirşof Devre Kanunları Mesh(Çevre Akımları) Analizi Node(Düğüm Gerilim) Analizi Thevenin Teoremi Norton Teoremi Maksimum Güç Transferi Yıldız Delta Dönüşümü Voltaj Kaynakları Akım Kaynakları Kirchhoff'un Gerilim Kanunu Kirchhoff'un Akım Kanunu Gerilim Bölücüler Akım Bölücüler Elektrik Enerjisi ve Güç