Learn New Ways of Current Divider Formula / Rule In 15 Minutes

For determining the total current through a resistor in a parallel circuit, it is required a powerful formula like the current divider formula. If the resistor is connected in series, the voltage across the load can be calculated by applying the voltage divider rule, which is a powerful voltage division rule when trying to find the voltage source across a resistor r in voltage divider circuits. Due to the fact that we are talking about series and parallel circuitry, letting me tell you a little about these two things. After that, we will discuss in detail the current divider formula and Kirchhoff’s current rule.

Series Resistors:

In order to understand the series connection, it is essential to understand that each resistor has two terminals to complete the electric circuit. The device is regarded as a two-terminal device because of the design of the connector. As shown in Figure, the resistor R2’s terminal 1 joins directly to the resistor R1’s terminal 2 on one side, and the remaining terminal 2 of the R2 is linked to the resistor R3’s terminal 1 on the other side. Since each resistor has only one connection, there would only be one.

current divider formula

Two terminal devices of resistor

Figure 1: Two terminals of a resistor

Series circuit of three resistors

Figure 2: Series connection of resistors

A series connection is stabilized when all the components i.e. all the resistors are joined in this way, this means that the resistors are linked physically together. In case three elements are connected to the same point, then the resistors R1 and R2 would not form a series network between each other. In essence, for resistors placed in series, the resistance of the entire series arrangement equal value of the sum of their individual resistance. Taking any number (N) of resistors as a whole, we have an equation.

R(T) = R1 + R2 + R3 + R4+ ………… + R(N) ………………………. Equation 1.

In a circuit with one or more resistors in series, the resistance of the circuit raises significantly regardless of the value of any resistor easily found as a consequence of Eq. (1). The resistance is determined largely by the characteristics of the largest resistor in the series combination. This is because the total resistance for the Fig. 2 configuration is

R(T) = R1 + R2 + R3=15 (KOhm) + 20 (KOhm) + 50 (KOhm)

= 85 (KOhm)

Calculation of Series Resistance:

Example 1: The series arrangement of a resistor on an ohmic system in series as shown in the below figure. If the output current reaches its complete destination at the load, find out the total resistance of the circuit. In this example of a network, all resistors have been matched to standard values.

 example of calculation of 4 resistors

Figure 3: Example of calculation of 4 resistors

Solution of this series circuit

It is important to note that the resistors’ series connections are kept as straightforward in Fig 3. Even though the resistor, R3, is on the vertical line, and the resistor at the bottom, R4, returns to the terminal. The following equations can be used to calculate the total resistor and these resistances have different values.

R(T) = R1 + R2 + R3 + R4=15 (KOhm) + 20 (KOhm) + 10 (KOhm) + 20 (KOhm)

R(T)= 65(KOhm)

When the resistors are the similar value as in equation (1), these modifications can be made easily, so long as the resistors remain the similar value.

R(T)=NR

A number of resistors N can be taken to be equal in value of R if they are connected in series.

SERIES CIRCUITS

2nd example: You can create a series circuit by connecting the voltage source in series with the series resistors in the figure in order to get the series circuit. Circuits are made up of various elements that are mingled together in a way that leads to a continuous flow of electrical charge, or current.

Second example of calculation series circuit

Figure 4: A complete circuit diagram of a series circuit

The first thing you need to understand is that a dc supply is also a two-terminal device, so you have to connect two points. We assume a series circuit has been established if we simply connect the supply at one end to the series of resistors. In order to determine which direction a conventional current flows, the supply’s connections must be properly arranged. For continuous dc circuits, the direction of the conventional current varies depending on the supply voltage. Given the positive voltage of a continuous dc supply, the direction of conventional current in a continuous dc circuit varies according to voltage. When it comes to analyzing series circuits it is important to keep in mind that the current flows in the same direction regardless of the elements that are connected together in series. Figure 4 illustrates a circuit with three resistors, during which current flows through them all at the same rate. Moreover, if the matter of whether two elements are arranged in series is ever a question, then you can look up whether the current is the same through both elements. No matter what the configuration is, the current must always be the same no matter how many elements are connected in series. The conditions in which two adjacent elements share the same current will not necessarily result in the elements being in series.

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In order to determine the current level and the voltage across each resistor, one must first ensure that the circuit is completed and the total circuit current has been determined. This is easily accomplished by using Ohm’s law and putting total resistance into the equation. That is,

Is=(E)/R(T)

Resistance of the circuit

Figure 5: Resistance of the circuit

Total Resistance of the circuit

Figure 6: Total resistance of the circuit

Figure 6 shows the total resistance to the circuit in Figure 4, resulting in the current I. For this configuration, the resultant current is I.

I=V/R(T)=I0/45=0.22Amp

There is the same value of the current Is at each corner of the network. The main point to be remembered here is that the power supply also indicates the current on the current display. We will now know the resistance value as well as the voltage across each resistor as we have the current level.

Understand that the direction of the current determines the polarity of the voltage across a resistor. Figure (4) illustrates the effect caused by current passing through a resistor that caused a voltage drop as shown in Figure (4). The polarity will be reversed with a reversal of the current direction.

It can be determined the magnitude of the voltage drop across each resistor using Ohm’s law, which is a mathematical rule, and when applied to the resistance of each resistor you can consider a small circuit with the resistor in series with the battery. That is,

We have already determined the current of the circuit as follows.

I=V/R(T)=I0/45=0.22mAmp

Since this is a series circuit, then the current of each path will be the same.  We can apply the below formula in order to determine the voltage of each resistor.

V1=I1*R1=Is*R1, V1=0.22*15K=3.3Kv

V2=I2*R2=Is*R2; V2=0.22*20K=4.4Kv

V3=I3*R3=Is*R3; V3=0.22*10 =2.2Kv

The result can be tabularized as follows:

Sl NoRIV
1R1=15K0.22A3.3Kv
2R2=20K0.22A4.4Kv
3R3=10K0.22A2.2Kv

Example of series circuit solution:

Next example of series circuit

Figure 7. Series Circuit Example

Find the total resistance RT in the series circuit shown in Fig. 15; find the resulting source current in and around the current-limiting resistor, and find the voltage across each resistor.

Solution:

R(T)=5+10+15=30(Ohm)

Current,

I = 5volts/30; I=0.16Amp

Since this is a series circuit, then the current of each path will be the same.  We can apply the below formula in order to determine the voltage of each resistor.

V1=I1*R1=Is*R1; V1=0.16*5 Ohm=0.8v

V2=I2*R2=Is*R2; V2=0.16*10=1.6V

V3=I3*R3=Is*R3; V3=0.16*15 =2.4v

The result can be tabularized as follows:

Sl NoRIV
1R1=5 Ohm0.16Amp0.8v
2R2=10 Ohm0.16Amp1.6v
3R3=15 Ohm0.16Amp2.4v

Parallel Resistors

Those of us who are familiar with the word parallel are aware that it refers to symmetry or similarity which occurs in the physical arrangement between two elements or two parallel branches. It is generally accepted that two elements, all the branches, or circuits that have two points in common are considered to be in parallel. For example, in Fig. 8(a), there are two resistors that are connected together at points a and b, forming parallel circuits. There is only one way to connect the resistors in parallel, and that is to connect both ends together as shown. Figure 8(b) shows two resistors are in parallel since points a and b are again in common. In R3 and R5 there is no parallelism because they are connected at a single point (b). Additionally, the connection between R3 and R5 is not in series, since a third connection (R4 resistor) is being found at point b. It is also noted that the same is true of the resistors R6 and R7. In addition to this, while the discussion above was written in relation to resistors, it can be applied to any kind of two-terminal device, whether it be a voltage source, meter, or any other device having two terminals.

Parallel Circuit

Figure 8: Parallel Circuit

Fig.9 illustrates a number of ways a parallel combination can appear on schematic parallel paths, as can be seen from the example shown in the previous part. There are three resistors connected as parallel resistors in each case. All of them have things in common with one another.

Different Types of Parallel Circuit

Figure 9: Different types of parallel circuit

With parallel resistors, as shown in Fig. 9, we can use the following equation to determine the whole resistance:

1/R(T) = 1/R1+1/R2+1/R3+……+1/R(N)

Due to the fact that G=1/R, the equation can be rewritten as follows, in terms of conductance levels:

G(T)=G1+G2+G3+……..+G(N)

N of Parallel Circuit

Figure 10: Parallel of number of resistance

Typically, however, the following format is used to get a whole resistance of, when the complete resistance is desired:

R(T) = 1/[1/R1+1/R2+1/R3+……+1/R(N)]

Practical Example of parallel circuit:

In Figure, determine the total resistance of the parallel network, separately calculate the total conductance of the parallel network, and finally figure out its relationship to the total resistance by using the circuit equation in parts (a) and (b).

Example 1

Figure 11: Example 1, calculation of parallel circuit

Here, we can determine the conductance then we can calculate the total resistance, and definitely, this will minimize the difficulty of the mathematical calculation.

G1=1/R1=1/5=0.2S

G2=1/R2=1/10=0.1S

Therefore, total conductance will be as follows-

G(T)=G1+G2; G(T)=0.2+0.1=0.3(S)

Hence,

R(T)=1/G(T)=1/0.3=3.33(Ohm)

We can also apply the below formula in order to determine the total resistance of the circuit.

R(T)=1/(1/R1+1/R2)=1/(1/5+1/10)=3.33(Ohm)

In general, no matter which method of resistance we adopt, the total resistance is the same as depicted above.

Example 2 of Parallel Circuit

Is it possible to define which parallel elements in Figure 12 have the least conductivity? Identify the total conductance of the network and make sure the conclusions on this are verified.

Example 2 of parallel circuit

Figure 12: Example 2, calculation of parallel circuit

On the above circuit, there are three different branches such as R1 is one branch, R2 is another branch resistances and R3 opposite branch resistance. By applying Eq. (3) and the results of part (a), determine the total resistance.

Solutions:

Because the 20 Ohm resistor stands for the most resistance and it will certainly have the lowest level of conductance (opposing the flow of charge), it will also have the highest resistance. So there will be the lowest individual branch currents.

G1=1/R1=1/5=0.2S

G2=1/R2=1/10=0.1S

G3=1/R3=1/20=0.05S

Therefore, total conductance will be as follows-

G(T)=G1+G2+G3; G(T)=0.2+0.1+0.05=0.35(S)

Hence,

R(T)=1/G(T)=1/0.35=2.85(Ohm)

We can also apply the below formula in order to determine the total resistance of the circuit.

R(T)=1/(1/R1+1/R2+1/R3)=1/(1/5+1/10+1/20)=2.85(Ohm)

Example 3 of Parallel Circuit

Example 3 of parallel circuit

Figure 13: Example 3, calculation of parallel circuit

In this example, the 13(a) circuit looks very complex, but when we rearrange the circuit as shown in figure 13(b), it looks very simple parallel circuit. Since all the individual resistances of the circuit are the same, we can use the below rule to determine the total resistance of the circuit.

R(T)=R/N=10K/4=2.5K Ohm

Parallel Circuits

In order to create a parallel connection, a power supply must now be connected across a parallel strip of resistors, as shown below One positive terminal from the power supply is connected directly to the positive terminal of the resistors. You always connect the negative terminal to the negative terminal. Due to the same applied voltage across both resistors, we can conclude that both resistors are in parallel with each other. The standard way to determine the voltages of parallel elements is to assume whatever voltage is present across all parallel circuits to be the same one.

In order for parallel connections to be useful, one has to be able to maintain the same voltage across the entire parallel connection. Additionally, Figure 14 also illustrates how the voltage along a circuit is affected by the voltages.

Parallel Network

Figure 14: Parallel Network

Battery Voltage=V1=V2

When the two parallel resistors of the supply are connected together, a supply current flows through the supply. It passes through each parallel resistor at the same time that the power is passing through the supply. In a circuit, the capacity of each element is proportional to its resistance, and the fractional currents flowing through each element can be calculated by calculating the resistance along the circuit. It is understood that the current maximum occurs when the resistance is smaller in an electrical circuit when it is designed along with series from the positive terminal to the negative terminal.

It is difficult to observe a parallel combination of elements on a serial circuit because they cannot be seen by the source within a serial circuit. Circuit resistance is very small compared with the total resistance. After determining current through Ohm’s law, you can use the data to determine the branch current for given loads.

Total Circuit of the Parallel Network

Figure 15: Total circuit of the parallel network.

Is=E/R(T)

Calculation of Parallel Circuit:

There are numerous parallel arrangements of elements that are difficult to observe on a serial circuit because they cannot be seen from the point of view of the source within a serial circuit. It is recommended that you check that equation (9) is satisfied in the parallel network.

Calculation of parallel circuit

Figure 16: calculation of parallel circuit

Solution:

Battery voltage = V1 = V2

Here, R(T)=(R1*R2)/(R1+R2)=10*20/(10+20)=6.67 Ohm

Utilizing Ohm law:

Is=V/R(T)=10/6.67

=1.5Ampere approximately.

So, V1=I1*R1

10=I1*10;I1=1amp

V2=I2*R2=I2*20

I2=10/20; I2=0.5Amp

We know Is=I1+I2

Is=1+0.5=1.5Amp (Proved)

Calculation 2 of the parallel circuit:

From the below circuit, we need to determine R3, and the current, and each resistor. Here, R(T) is given as 10 Ohm.

Example 2 of parallel circuit

Figure 17 Calculation 2 of parallel configuration circuit

We know,

R(T)=1/(1/R1+1/R2+1/R3)

Putting the given value of the resistor, we get:

30=1/(1/10+1/20+1/R3); 1/10+1/20+1/R3=10; 0.1+0.05-0.1=1/R3

R3=29.85; R3=29.85 Ohm

Now we can able to determine the current using Ohm law.

V1=R1*I1;I1=10/10; I1=1Amp

Similarly,

I2=10/20; I2=0.5Amp

I3=10/29.85; I3=0.33Amp

The total current entering=I1+i2+I3

total current flowing =1+0.5+0.33=1.83Amp.

Parallel Circuit Current divider formula

Let us now introduce a straightforward methodology to find the current flowing through a resistor inside a parallel circuit utilizing the current divider rule (CDR). A current divider formula is a straightforward method that can be applied to many circuits with resistances of a given value.

As the current seeks the path of least resistance from all the connected paths, so the current will flow the lowest resistance r1. To illustrate this point, let us consider the case of being faced with a current of 9 A divided between three parallel resistors. 

Current Divider

Figure 18  circuit

The 10 Ohm resistor r1 will undergo the greatest amount of current flow, while the 40 Ohm resistor r2 will undergo the smallest amount. Based on what, we have already discussed. As discussed above, therefore, the difference between the current flowing through the 100 Ohm resistor and the 1K resistor is that the 100 resistor has a much higher current flowing through it. It is important to recognize that the resistance of the 100 Ohm resistor has 10 times the resistance of the 10 Ohm resistor so that we can take it one step further.

There is a result of a current running through a 10 Ohm resistor that is 10 times higher than the current running through a 100 Ohm resistor. The 100-ohm resistor has a 10x current than the one k ohm resistor.

A current divides / divide equally between two elements of the same value if they are parallel. There is a direct correlation between the resistance values of parallel elements, and the input current, in terms of how much current is required for each of them independently. The current from the current source splits with a ratio equal to the inversely proportional or inverse proportion of the resistance values of the elements between parallel elements. For parallel elements that have the same resistance, the current from the current sources split with the same ratio of their resistance values.

Example of current divider rule:

Example of Current Divider circuit

Figure 19: Example of a parallel circuit

In the above circuit, R1, R2 and R3 are connected in parallel. Here R1 and R2 are two parallel resistors with R3 and I2 is given as 2Amp, therefore, we need to find out the I1, I3, and Is.

Since we are given I2, therefore, the voltage of R2 would be as follows-

V2=20*2 =40Volts

So, I1=40/10; I1=4Amp; I3=40/40; I3=1Amp

Therefore, calculate the current would be as follows:

Is=I1+I2+I3; Is=4+2+1; Is=7Amp

Current Divider Rule for a circuit

Figure 20: Parallel Circuit Digram

The diagram above illustrates a fairly straightforward resistive current divided circuit that can be seen as a simple circuit, but when shown in this manner, you will be able to see more about how it works. Due to the fact that the voltage V is constant across all parallel elements, the following is true as per general formula:

V-I1*R1-I2*R2-I3*R3-………………..-Ix*Rx

The equation above is converted into the following when V is replaced for IT.

IT=(Ix*Rx)/RT

When Ix is subtracted, the resulting current division rule is the following one:

Ix=(RT*IT)/Rx

Which states that

Even though this circuit is a simple one, it is clear that you will be able to see how a current divider circuit, which is one kind of linear circuit behaves when it is demonstrated in the above way.

Examples Current Divider Formula:

It is easy for us to understand the concept of the current divider circuit by referring to the practical example. I have included a circuit diagram to demonstrate resistive current divider circuits using parallel circuits demonstrating the principle of current dividing. It is important to note that the following calculation needs to be performed and I have calculated I1, I2, and I3, using the current divider rule, for the circuit below.

Example of current divider formula

Figure 21: Example of current divider formula

Solution:

To determine the current, first, we need to find out the total equivalent resistance of the current divider circuit. And, we know the following parallel connection formula in order to find out the equivalent resistance for the current divider circuit.

R(T)=1/(1/R1+1/R2+1/R3); R(T)=1/(1/5+1/10+1/15)

=1/(0.2+0.1+0.067); R(T)=1/0.367

R(T)=1/0.367; R(T)=2.72 Ohm

Now, we know

I1=RT*IT/R1; I1=2.72*30/5; I1=16.32Amp

I2=2.72*30/10; I2=8.16Amp; I3=2.72*30/15; I3=5.44Amp

Justify

In general case, we know

IT=I1+I2+I3; =16.32+8.16+5.44=30Amp

The result is shown in the below table after applying the current divider formula.

Sl NoRTRITIJustify
12.72 OhmR1=5 OhmIT=30AmpI1=RT*IT/R1 I1=16.32AmpIT=I1+I2+I3 =16.32+8.16 +5.44 =30Amp    
22.72 OhmR2= 10 OhmIT=30AmpI2=RT*IT/R2 I2=8.16Amp
32.72 OhmR3= 15 OhmIT=30AmpI3=RT*IT/R3 I3=5.44Amp

In a short circuit, the resistance equal/ r total to zero means the entire current flows.

Summary

In particular, on other words, it is important to know that the current divider rule is one of the most important and widely used rules in electrical engineering. As an engineer, you should always know Kirchhoff’s laws because it can help you make quick and accurate decisions when designing electric circuits in terms of the current divider circuit. If you are a beginner and are not aware of the current divider rule, read this article now and learn the main benefits of using this rule.