If the resistor combination provides multiple paths for the flow of current, it is called resistors in parallel. The equivalent resistance always decreases when we add parallel resistors. Use the online parallel resistors calculator here.

## Parallel Combination:

Resistor combination would be called **resistors in parallel** if there are more than two components are connected to the same node in a circuit.

Suppose we have three resistors connected to each other at point b, the head of the first resistor is connected to the head of the second and third resistor. Assuming that the current is flowing from top to bottom. In contrast to the series combination, the positive terminal of R_{1} is connected to the positive terminal of R_{2 }and R_{3}. The combination will lead to three different paths for current.

Notice that there are four components present at node a and node b; the source, R_{1}, R_{2,} and R_{3}, as the diagram shows.

*The parallel circuit provides more than one path for current.*

## Online Parallel Resistors Calculator:

To calculate the total resistance of the resistors, which are connected in parallel, use the following parallel resistors calculator. $R_1, R_2$, and $R_3$ are the three resistors connected in parallel and the $R_{Total}$ is the equivalent resistance.

## Parallel Circuit Current Calculator

After calculating the total resistance of parallel resistors, now you can calculate the total current. Put the total resistance from the above online parallel resistors calculator. Also, input the circuit voltage to calculate the current.

## Solving Resistors in Parallel circuit:

We want to know the few types of the parameter for the above circuit as we did for the series combination.

### Total Equivalent Resistance:

##### What is the total current provided by the source?

To answer the question, we need to calculate the total or equivalent resistance of the circuit. And the formula for that is:

$\frac{1}{R_{Equivalent}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\ldots +\frac{1}{R_{n}}$

If we use the same resistor values, we used in the series circuit i.e. R$_{1}$, R$_{2,}$ and R$_{3}$ are 20, 40, and 10 $\Omega$ respectively. And put these values in the above formula, then:

$\frac{1}{R_{Eq}}=\frac{1}{20}+\frac{1}{40}+\frac{1}{10} =\frac{7}{40}$

$R_{Eq}=\frac{40}{7}=5.714 \Omega $

The above complex circuit is now reduced to this single resistor circuit. Now we can easily find out the total current the circuit is consuming by ohm law as shown below:

$I_{T}=\frac{V}{R_{Eq}}=\frac{140 v}{5.714 \Omega }$

$I_{T}=24.5 A$

*The total resistance is far less than that for the series and the total current far greater than that for the series circuit.*

### The voltage Across Each Resistor:

##### What is the voltage across each resistor?

First of all, take a look at the circuit and you will realize that each and every resistor is connected to a voltage source. The voltage across each resistor is the same as the source voltage. The formula for voltage is:

$V_{T}=V_{1}=V_{2}=\ldots =V_{n} $

In our case, voltage for R$_{1}$, R$_{2}$ and R$_{3 }$are: V$_{1 }$= 140 v, V$_{2 }$= 140 v, and V$_{3 }$= 140 v respectively.

### The current of Each Resistor:

##### What is the current of each resistor?

We can use Ohm’s law for finding the current. Suppose that the current for R$_{1}$, R$_{2,}$ and R$_{3 }$are I$_{1}$, I$_{2,}$ and I$_{3 }$ respectively. So, by applying Ohm’s law over each resistor we get

$I_{1}=\frac{V_{1}}{R_{1}}=\frac{140 v}{20 \Omega }=7 A$

$I_{2}=\frac{V_{2}}{R_{2}}=\frac{140 v}{40 \Omega }=3.5 A$

$I_{3}=\frac{V_{3}}{R_{3}}=\frac{140 v}{10 \Omega }= 14 A$

Now, what if we add all these individual currents? Let’s do this.

$I_{T}=I_{1}+I_{2}+I_{3} $

$I_{T}=7 A+3.5 A+14 A$

$I_{T}=24.5 A$

The same current be can be confirmed from the parallel resistors calculator above.

Finally! We got the total current drained by all these resistors as we found the first place. It means that the total current of the circuit is the same as the summation of individual resistor currents. We can generalize the formula like this:

$I_{T}=I_{1}+I_{2}+\ldots +I_{n}$

The individual current of each resistor can be found through the Current Divider Rule (CDR).

## The formula to Add Parallel Resistors:

We have observed three parameters in the above discussion: First total resistance in parallel, second, the individual voltage of each resistor, and lastly, the individual current of each resistor. All these formulas are presented here for your simplicity with generic applicability for up to n number of resistors.

The total resistance of the parallel circuit:

$\frac{1}{R_{Equivalent}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\ldots +\frac{1}{R_{n}}$

The individual voltage of resistors is the same as the source voltage:

$V_{T}=V_{1}=V_{2}=\ldots =V_{n}$

The total current is the addition of all individual currents:

$I_{T}=I_{1}+I_{2}+\ldots +I_{n}$

## Add Parallel Resistors of Same Value:

Suppose a circuit where all the resistor has the same value and all of them are connected in parallel, what will be the total resistance?

Let us assume that $n$ number of resistors in parallel with having the same resistance. The total resistance of the circuit will be

$R_{Total} = \frac{R}{n}$

The $R$ is the resistance value, which is the same for all the resistors and the $n$ is the number of resistors connected in parallel.

#### What will be the current in each resistor?

The current of each resistor will

$I = \frac{I_{Total}}{n}$

## Conclusion:

- Current divides in a parallel circuit
- Add parallel resistors to increase the paths for current
- The total current is equal to the individual current of parallel paths
- Voltage remains the same in parallel paths of an electrical circuit
- The total resistance reduces in a parallel circuit as compare to individual parallel path resistance

Please how did you get the 7/40, before you divided it and get R equivalent

It is a basic Math problem of the fraction addition. For making the denominators of all fraction same, you have to multiply and divide it by a number. The above will become

(2*1)/(2*20)+1/40+(4*1/4*10)

=2/40+1/40+4/40

=7/4