# Current Divider Rule (CDR) – Solved Examples for AC and DC Circuits

## Current Division “CDR” for for Resistive, Inductive and Capacitive Circuits

**What**** is Current Divider Rule (CDR)?**

When a number of elements are connected in parallel, the current divides into a number of parallel paths. And the voltage is the same for all elements which are equal to the source voltage.

In other words, when the current passes through more than one parallel path (the voltage divider rule “VDR” or voltage division is used to calculate the voltage in the series circuits), the current divide in each path. The value of current passes through a particular branch depends on the impedance of that branch.

The current divider rule or current division rule is the most important formula that is widely used to solve circuits. We can find the current that passes through each branch if we know the impedance of each branch and the total current.

The current always flows through the least impedance. So, the current has an inverse relationship with impedance. According to ohm’s law, the current that enters the node will be split between them in inverse proportion to the impedance.

It means that the smaller value impedance has a larger current as the current chose the least resistance path. And the larger value resistance has the least current.

According to the circuit elements, the current divider rule may describe resistors, inductors, and capacitors.

Related Posts:

*Current Divider Rule Calculator – CDR Formula & Calculations*

*Voltage & Current Divider Rules (VDR & CDR) Equations*

**Current Divider Rule for Resistive Circuits**

To understand the resistive current divider rule, let’s take a circuit in which the resistors are connected in parallel. The circuit diagram is shown in the figure below.

In this example, a DC source supply to all resisters. The voltage of resisters is the same as the source voltage. But due to parallel connection, the current divides into different paths. The current divides at each node and the value of current depend on the resistance.

We can directly find the value of current passing through each resistor with the help of the current divider rule.

In this example, the main current supplied by the source is I. And it divides into two resistors R_{1} and R_{2}. The current passes through the resistor R_{1} is I_{1} and the current passes through the resistor R_{2} is I_{2}.

As the resistors are connected in parallel. So, the equivalent resistance is R_{eq}.

Now, according to Ohm’s law;

*V = I R _{eq}*

Both resisters are connected in parallel with a DC source. Therefore, the voltage across the resistor is the same as the source voltage. And the current that passes through the resistor R_{1} is I_{1}.

So, for resister R_{1};

Similarly, for resister R_{2};

So, these equation shows a current divider rule for resistance connected in parallel. From these equations, we can say that the current that passes through resister is equal to the ratio of multiplication of total current and opposite resistance with the total resistance.

*Related Posts: *

*Thevenin’s Theorem. Step by Step Guide with Solved Example**Norton’s Theorem. Step by Step Guide with Solved Example*

**Current Divider Rule for Inductive Circuits**

When inductors are connected in parallel, we can apply the current divider rule to find the current that passes through each inductor. To understand the current divider rule, we take a circuit in which the inductors are connected in parallel as shown in the figure below.

Here, two inductors (L_{1} and L_{2}) are connected in parallel with a source voltage V. Total current passes through the source is I ampere. The current passes through the inductor L_{1} is I_{1} and the current passes through the inductor L_{2} is I_{2}.

Now, we need to find the equations for current I_{1} and I_{2}. For that, we will find the equivalent inductance L_{eq};

We know that the total current passes through the circuit are I and it is equating as;

So,

Now, for inductor L_{1}, current passes through this inductor is I_{1};

For inductor L_{2};

The current divider rule for the inductor is the same as the current divider rule for the resistors.

Related Posts:

*Superposition Theorem – Circuit Analysis with Solved Example**Maximum Power Transfer Theorem for AC & DC Circuits*

**Current Divider Rule for Capacitive Circuits**

When the capacitors are connected in parallel, we can find the current passes through each capacitor by using the current divider rule. To understand the current divider rule for the capacitor, we take an example in which the capacitors are connected in parallel as shown in the figure below.

Here, two capacitors (C_{1} and C_{2}) are connected in parallel with a voltage source V. The current passes through the capacitor C_{1} is I_{1,} and the current passes through the capacitor C_{2} is I_{2}. The total current supplied through the source is I.

Now, we need to find the equations for current I_{1} and I_{2}. For that, we will find the equivalent capacitance C_{eq};

*C _{eq} = C*

_{1}

*+ C*

_{2}

We know the equation for the current that passes through the capacitor. And the equation for the total current supplied by the source is;

For capacitor C_{1}, the current that passes through this capacitor is I_{1};

For Capacitor C_{2};

The current divider rule for the capacitor is slightly different from the current divider rule for the inductor and resistor.

In the capacitor current divider rule, the current passes through a capacitor is a ratio of the total current multiplied by that capacitor to the total capacitance.

Related Posts:

*Millman’s Theorem – Analyzing AC & DC Circuits – Examples**Tellegen’s Theorem – Solved Examples & MATLAB Simulation*

### Solved Examples for AC and DC Circuits using CDR

#### Current Diver Rule for DC Circuit

**Example:1**

Find the current passes through each resistor by the current divider rule for the given network.

In this example, three resistors are connected in parallel. First, we find the equivalent resistance.

*R _{eq} =* 100/17

*R _{eq} =* 5.882 Ω

The total current supplied by the source is I. So, according to ohm’s law;

*V = I R _{eq}*

50V = *I* (5.882Ω)

*I = *50V / 5.882Ω

*I = *8.5* A*

Now, we apply the current divider rule to the first resister (10 Ω), and the current passes through this resister is I_{1};

Here R_{2} and R_{3} are connected in parallel. So, we need to find the equivalent resistance between R_{2} and R_{3}.

(*R*_{2} || *R*_{3} ) = 14.285 Ω

*I*_{1} = 4.9999 ≈ 5 A

Similarly, we apply the current divider rule to the Second resistor (20 Ω), and the current that passes through this resister is I_{2};

Here,

(*R*_{1} || *R*_{3} ) = 8.33 Ω

*I*_{2} = 2.499 ≈ 2.5 A

Now, we apply the current divider rule to the third resistor (50 Ω), and the current that passes through this resistor is I_{3}.

Here,

(*R*_{1} || *R*_{2} ) = 6.66 Ω

*I*_{3} = 1.00 A

So, the summation of all three currents will be;

*I*_{1} + *I*_{2} + *I*_{3} = 5 + 2.5 + 1 = 8.5 A

And this current is the same as the total current supplied by the source.

Related Posts:

*Voltage Divider Rule (VDR) – Solved Examples for R, L and C Circuits**Voltage Divider “VDR” Calculator, Examples & Applications*

#### Current Diver Rule for AC Circuit

**Example-2**

Consider an AC circuit having a resistor and capacitor connected in parallel as shown in the figure below. Find the current passes through the resistor and capacitor using the current divider rule. Consider 60 Hz frequency.

Z_{R} = 200 Ω = 200∠0°Ω

Z_{C} = 1/(2 π*f*C) = 1/(2 π 60(5×10^{6}) )

Z_{C} = 10^{6} / (600 π)

Z_{C} = 530.78 ∠-90° Ω

Now, according to the current divider rule, the equation of current passes through the resistor is;

Now, similarly, we can find the current passes through the capacitor. According to the current divider rule, the equation of current passes through the capacitor is;

*I _{C} *= 120 ∠0° (0.3526 ∠ 69.353°)

*I _{C}* = 42.31 ∠ 69.353°

If you want to prove this answer, you can add both currents. And the value of this current is the same as the source current.

**Related Electric Circuit Analysis Tutorials:**

- Compensation Theorem – Proof, Explanation and Solved Examples
- Substitution Theorem – Step by Step Guide with Solved Example
- SUPERNODE Circuit Analysis – Step by Step with Solved Example
- SUPERMESH Circuit Analysis – Step by Step with Solved Example
- Kirchhoff’s Current & Voltage Law (KCL & KVL) | Solved Example
- Cramer’s Rule Calculator – 2 and 3 Equations System for Electric Circuits
- Wheatstone Bridge – Circuit, Working, Derivation and Applications
- Electrical and Electronics Engineering Calculators
- 5000+ Electrical and Electronic Engineering Formulas & Equations