Current Divider Rule

To understand this rule, one needs to have an idea about electric circuits and their classification according to the arrangement of components in the given circuit. Because as per their working, they act as divider circuits.

Table of Content

• What is the Current Divider Rule?
• Derivation of Current Divider Rule
• Important Conditions of Current Divider Rule
• Sample Problems on Current Divider Rule
• When to Use the Current Divider Rule?
• Advantages of Current Divider Rule
• Disadvantages of Current Divider Rule
• Applications of Current Divider Rule

Electric Circuit

An electric circuit provides a path through which electric current can be transmitted. It includes a device like a battery or a generator that provides energy to the charged particles constituting the current. To understand the current divider rule, will discuss its classification.

Types of Electric Circuit

Electric circuits are of two types named series circuits and parallel circuits according to the arrangement of components.

1. Series Circuits
2. Parallel Circuits

• Series Circuits: It is a kind of circuit where the components are connected in a chain. It acts as a voltage divider rule.
• Parallel Circuits: It is a kind of circuit where the components are connected between two common points. It acts as a current divider rule.

What is the Current Divider Rule?

According to this rule, the current in each branch of a given circuit is represented as a portion of the total current flowing in the circuit.

The generalized equation of the current divider rule is:

IBranch  =  ITotal * ()
WHERE:
IBranch = Current flowing through the given branch
ITotal     =  The total current flowing through the circuit
Req      =  The equivalent resistance of the circuit
RBranch=  The total resistance in a given branch of the circuit 

Derivation of current divider rule

Consider the following circuit:

According to ohms law,

V = I_T * R_eq ┉┉┉┉ ➀

As the resistors are in parallel the voltage across each of them is ‘ V ‘

So, according to ohms law,

V = I₁R₁ ┉┉┉┉ ➁

V = I₂R₂ ┉┉┉┉ ➂

Substitute ➁ in ➀

V = I₁R₁ = I_T * R_eq

⇒I₁ = I_T * ()

We know,

I_T = I₁ + I₂

⇒I₂ = I_T – I₁

I₂ = I_T – [I_T * ()]

Upon solving this we get,

I₂ = I_T * ()

Derivation using Conductance

The conductance of any element of a circuit is mathematically given as ‘(1/Resistance)’.

Consider the same circuit as above where

V_T = I_T * R_eq

V_T = I_T * () ┉┉┉┉ ➀

As the resistors are in parallel the voltage across each of them is ‘ V ‘

So, according to ohms law,

V = I₁R₁ = ┉┉┉┉ ➁

V = I₂R₂ = ┉┉┉┉ ➂

Substitute ➁ in ➀ and ➂ in ➀,

= I_T * ()

⇒I₁ = I_T * ()

= I_T * ()

⇒I₂ = I_T * ()

Important conditions of Current Divider Rule

Note that there are certain constraints under which the above formula for current divider rule is derived.

These are as follows:

• All the sources in the circuit whether it be a current source or a voltage source, are considered being ‘ideal’.
  This means that the internal resistance in the source elements is considered to zero.
• All the elements in the circuit are considered being ‘linear’ so their properties won’t change with respect to time.

Sample Problems on Current Divider Rule

Problem 1:

Calculate the magnitude of current in each branch of a circuit consisting of two resistors of resistance 20Ω and 30Ω connected in parallel combination. Input voltage is given as 60V.

Solution:

To apply the current divider rule, we need to calculate the total current and equivalent resistance first.

Equivalent Resistance:

R_eq =

R_eq =

=
= 12 Ω

Total current I_T =

I_T =

I_T = 5A

By current divider rule:

I₁ = I_T * ()

⇒I₁ = 5 * () = 3A

⇒I₂ = 5A – 3A = 2A

Problem 2

In the following circuit, calculate the magnitude of I₂. The input voltage is 30 volts.

Solution:

Equivalent resistance:
R_eq = R₁ + R₂ || R₃

R_eq = 10 + = 15Ω

Total Current:

I_T = = = 2A

Method 1: Without using the current divider rule

The voltage across R1 :

V1 = I_T * R1 = 2 * 10 =20V

⇒Voltage across R2 and R3 is V_input – V1 = 30 – 20 =10V

Current across R3 i.e I2 = V_R3/R3= 10/10 = 1A

Method 2: Using the current divider rule

Let the equivalent resistance up to the point where the total current starts dividing is ‘K’

K= R₂ *

K =

K = 5Ω

By current divide rule:

I₂ = I_T * (K/R₃)

I₂ = 2 * (5/10) = 1A

When to use current divider rule?

We can use the current divider rule in the following conditions:

• Parallel Circuit Configuration
• When parallel branches is connected to same circuit source.
• The branches must have resistive elements. It should not contain any impedance components.

Advantages of Current Divider Rule

• It is easy to apply when the circuit will have parallel branches. 
• It makes the calculation easier for network analysis. 
• It provides an intuitive understanding of the division of current in the branches.

Disadvantages of the Current Divider Rule

• The current divider rule assumes that the circuit is ideal i.e. purely resistive sources. It does not consider parasitic capacitance or inductance which can affect the current distribution. 
• When resistances are significantly different in magnitude, then the rule may lead to an imbalanced current division leading to errors in the calculation.
• It assumes the voltage source to be constant. If voltage fluctuates or if there are multiple voltage sources, it might lead to errors.

Applications of Current Divider Rule

• Current division is used to calculate the current in the branches of a circuit in which resistors are in parallel.
• It is used in making LEDs as the limiting current required in each branch for the filament to glow can be known using this rule.
• Used in charging and discharging of batteries in which current flow to each battery can be controlled using the current divider rule.
• A current also acts as a signal in many electronic circuits, this rule helps in dividing the signal into multiple channels along the communication pathway.

Conclusion

The current divider rule has wide range of day to day applications in communication systems, working of batteries, and even in the glowing of LED bulbs. It is used in finding the magnitude of current flowing in a particular branch of the circuit. It is mathematically given as

IBranch = ITotal * (\frac{R_{eq}}{R_{branch}})

Also note that this formula works only under the conditions that all sources in a given circuit are ‘ideal’ and all the elements in the circuit are ‘linear’ in nature.

FAQs on Current Divider Rule

Q.1: How would the formula change if in a given circuit capacitors and inductors were also present?

Answer:

If capacitors and inductors are also in the circuit, then the equivalent resistance will actually be net impedance of the circuit. Impedance is denoted by Z and is is given as:

Z = √[(R_eq)²+(Xc – X_L )²]

Here, the formula is given as follows:

I_Branch = I_Total * [Z_Total/Z_Branch] where,

Z_Total = Total impedance of the circuit
Z_Branch = Impedance of given branch

Q.2: What assumptions were made about a given circuit during the derivation of the current divider formula?

Answer:

The following assumptions are made:

(i) All the sources in the circuit whether it be a current source or a voltage source, are considered being ‘ideal’.
This means that the internal resistance in the source elements is considered being zero.

(ii) All the elements in the circuit are considered being ‘linear’ so their properties won’t change with respect t0 time.

Q.3: How to apply the current divider rule in case of complex circuits, like bridges?

Answer:

Application of current divider rule is same as done in above sample problems for any type of circuit.

In case of complex circuits, like bridges, other techniques like superposition theorem, voltage divider rule etc might be needed to be applied in order to find the current of a particular branch.