## 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 R_{eq} will be:

*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 V_{1} is the voltage drop of 1^{st} resistor, V_{2} is the voltage drop of 2^{nd} resistor and V_{n} is the voltage drop of n^{th} 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 I_{1} is the current through the 1^{st} resistor, I_{2} is the current through the 2^{nd} resistor and I_{n} is the current through the n^{th} 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.

Load this circuit in Simulator

*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.

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