## Resistors in Parallel

### Designing with resistors in parallel

Similarly, Let's see what happen in current and voltage when the resistors are connected in parallel. How to calculate total resistance if n number of resistors are connected in parallel as shown in figure.

The reciprocal $\mathsf{\frac{1}{R_{eq}} }$value is equal to the sum of reciprocal of individual resistance values.

\mathsf{ \begin{align} \frac{1}{R_{eq}} &= \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \; ... \; + \frac{1}{R_{n}} \\ (or) \\ \frac{1}{R_{eq}} &= \sum_{i=1}^n \frac{1}{R_i} \end{align} }

The Total resistance Req will be:

\mathsf{ \begin{align} R_{eq} &= \frac{1}{\sum_{i=1}^n \frac{1}{R_i}} \end{align} }

#### Voltage

The total voltage that appears across a parallel resistors network is the same as the voltage drops at each individual resistance.

Total voltage that appears across the circuit:

$$\mathsf{V\:\:=\:\:V_{1}\:=\:V_{2}\:=\:V_{3}\:=\;...\;\:=\:V_{n}}$$

Where V1 is the voltage drop of 1st resistor, V2 is the voltage drop of 2nd resistor and Vn is the voltage drop of nth resistor in the above circuit.

#### Current

The total amount of current entering a parallel resistive circuit is the sum of all individual currents flowing in all the Parallel branches. The resistance value of each branch determines the value of current that flows through it.

Current through the circuit:

$$\mathsf{I\:\:=\:\:I_{1}\:+\:I_{2}\:+\:I_{3}\:+\;...\;\:+\:I_{n} }$$

Where I1 is the current through the 1st resistor, I2 is the current through the 2nd resistor and In is the current through the nth resistor in the above circuit.

#### Example circuit

Let's take a example circuit, 3 resistors have resistance value 2 ohm, 4 ohm, and 8 ohm respectively are connected as parallel. This circuit shown in below.

1. Find the Current value through each resistor and the total current through this circuit, If Vin equal to 5V.

First, we need to find total resistance, As we know in parallel circuit, the reciprocal $\mathsf{\frac{1}{R_{eq}} }$value is equal to the sum of reciprocal of individual resistance values. i.e,

\mathsf{\begin{align} \frac{1}{R} &= \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \\ \frac{1}{R} &= \frac{ 4 + 2 + 1}{8} \\ \frac{1}{R} &= \frac{7}{8} \\ R &= \frac{8}{7} = 1.143\Omega. \end{align}}

From Ohm's Law, $\mathsf{ I = \frac{V}{R}}$.

So the total current,

$$\mathsf{ I = \frac{5}{1.143} \;= \; 4.37 A. }$$

The Current value through 2Ω resistor is:

Again from Ohm's Law $\mathsf{ I = \frac{V}{R}}$.

$$\mathsf{ I = \frac{5}{2} \;= \; 2.5 A. }$$

Similarly,

The Current value through 4Ω resistor is $\mathsf{ \frac{5}{4} \;= \; 1.25 A}$.

The Current value through 8Ω resistor is $\mathsf{ \frac{5}{8} \;= \; 0.625 A}$.

The total resistance value of the resistors which are in parallel would be less than the minimum resistance value of the resistor is in that circuit.