If the resistor combination provides multiple paths for the flow of current, it is called resistors in parallel. The equivalent resistance always decreases with an addition of resistor element.

## 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 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 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 the 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 paths for current.*

## Resistors in parallel calculator:

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

## 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 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 reduce 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 series and 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 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} $

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

$I_{T}=24.5 A$

Finally! We got the total current drained by all these resistors as we found the first place. It means that total current of the circuit is 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 Current Divider Rule (CDR). \(\)

## Resistor in Parallel Formula:

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 are same as source voltage:

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

The total current is the addition of all individual currents:

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

## Same value resistors in parallel:

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

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

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

The $R$ is the resistance value, which is 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
- The total current is equal to the individual current of parallel paths
- Voltage remains 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