Current Division Rule:

Current Divider Rule is a renowned method of solving parallel circuits. You can solve parallel circuits using Ohm's law. But Ohm's law has a limitation. You have to know the voltage across parallel elements. This method allows you to find out the current flowing through parallel elements without knowing the voltage across it.  This procedure is derived from two very popular circuit solving laws, is Ohm's law and KCL. In this article, I am going to derive the expression of the current divider rule as well. 

Explanation:

It is only acceptable to parallel circuits where the voltage remains the same throughout the circuit. Consider a parallel circuit given below.

Figure 1: A parallel circuit

Let,

i1 = Current across R1

i2 = Current across R2

i3 = Current across R3

IT = total current across the circuit

V1 = voltage across each element

Apply KCL,

IT = i1 + i2 + i3  ….. Equation 1

From Ohm's law, i = v/R, replace "i" in Equation 1.

IT = V1/ R1 + V1/R2 +V1/R3 

IT = V1 (1/R1 + 1/R2 + 1/R3)

As you know formula for parallel resistances is

1/RT = 1/R1+1/R2+1/R3

RT = R1 || R2 || R3

IT = V1 /RT 

IT = V1/RT..... Equation 2

Now, if you want to find the current across R2, then using Ohm's law:

i2 = V1/R2 …... Equation 3

Now, solve Equation 2 and Equation 3 to get the value of i2 which is independent of V1.

IT*RT = i2*R2

Or

i2 = IT*RT/R2

Similarly, 

i1 = IT*RT/R1

i2 = IT*RT/R2

i3 = IT*RT/R3

These are the equations of current in parallel circuits which are independent of voltage. It is only applicable to parallel circuits only. These types of circuits are also called Current Divider Circuits because of the current divide among all the resistances. As you know the current adopts the least resistive path. So, the lower the resistance the higher the current flows through it.

Solved Example:

In the previous article (parallel resistance formula), I analysed and solved a parallel circuit using Ohm's law. In that case I use the following circuit.

Figure 2: Parallel circuit with a voltage source

In this tutorial, I am going to solve the same examples with the help of the current divider rule. In this example I am going to use the circuit in figure 3. Both circuits (Circuits in figure 2 and figure 3) are the same irrespective of the current and voltage sources. In the later circuit a voltage source is replaced by the equivalent current source. 

Figure 3: Parallel circuit with an equivalent current source

Example 1:

Determine total or equivalent resistance and the current flows through each resistor with the help of the current divider rule.

I solved this problem in my previous article (parallel resistance formula).

This is another way to solve this problem.

Find total resistance RT

RT = 1/R1 + 1/R2 + 1/R3

RT = 545.5 Ω

Apply CDR on each resistor.

Current through R1 is i1,

i1 = IT*RT/R1

i1 = 18*545/1000

i1 = 9.8 mA

Current through R2 is i2,

i2 = IT*RT/R2

i2 = 18*545/2000

i2 = 4.9 mA

Current through R3 is i3,

i3 = IT*RT/R3

i3 = 18*545/3000

i3 = 3.2 mA

Conclusion:

Subsequently reviewing the current divider rule, its derivation and a solved example, it is deduced that:

• This technique is useful in finding the current flows through the resistors without knowing the voltage

• The lesser the resistance the larger the current 

• It is only acceptable to parallel circuits only