# Explain The Current Divider Rule (CDR) Parallel electrical circuits, the current doesn’t remain the same. The current divider rule(CDR) is used to find the divided current in parallel circuits.

## Statement  of Current Divider rule:

Statement: The electrical current entering the node of a parallel circuit is divided into the branches. Current divider formula is engaged to calculate the magnitude of divided current in the circuits.

#### Current Divider Formula:

A common formula for the current IX in a resistor RX that is in parallel with a combination of other resistors of total resistance RT is : Where IT is the total current entered into the network of RX in parallel with RT.  when RT is composed of a parallel combination of resistors, say R1R2, …… then the mutual of each resistor must be added to find the total resistance RT:  Fig:1

#### Derivation:

A normal circuit with two resistors in parallel with a voltage source is called a current divider. The current I has been divided into Iand I2 into two parallel branches with the resistance Rand R2 and V is the voltage drop across the resistance R1 and R2.

As we know

V = IR ……..(1)

Equation of the current is written as,

#### Let the total resistance of the circuit be R and is given by the equation shown below,

#### Equation (1) can also be written as

I = V/R ……….(3)

Now, putting the value of R from the equation (2) in the equation (3) we will get here,

#### Putting the value of V = I1R1 from the equation (5) in the equation (4), we finally get the equation as And now considering V = I2R2 the equation will be,

#### Thus, from the equation (6) and (7) the value of the current I1 and I2respectively is given by the equation below,

#### Thus, In the current division rule, It is said that the current in any of the parallel branches is equal to the proportion of opposite branch resistance to the total resistance, multiplied by the total current.

#### Example of Current Divider:

An electrical circuit has two parallel resistors of 2 ohms and 10 ohms. Apply the current divider equation to find the current flowing through both resistors when the input is 5 Ampere.

Solution: Similarly, we can solve the three parallel resistors.

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2. Pieter Beerten says:

Thanks for the nice explantion ! There is however a small mistake in equation 7. It says I = I1*R1/(R1+R1) where it should say I = I2*…

Anyways, well done!

1. Pieter Beerten says:

Off course, the division and multiplication should be switched in my comment, but you get the point!

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