Current Divider: What is it? Formula, Rule & Examples

Contents

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Key learnings:

Current Divider Definition: A current divider is defined as a circuit where the input current splits among multiple parallel paths according to specific ratios determined by the components’ resistances.
Formula Application: To calculate the current through any branch in a parallel circuit, divide the total circuit current by the resistance of the branch, then multiply by the total resistance of the circuit.
Derivation Insight: Understanding the derivation helps predict how currents distribute in complex parallel circuits, reinforcing the current divider rule’s application.
Practical Examples: Examples in the article illustrate how the current divider rule applies to different scenarios, aiding practical understanding.
Rule Utilization: The current divider rule is essential when dealing with parallel circuits to determine individual branch currents effectively.

What is a Current Divider?

A current divider is a circuit in which the output current is a share of the input current. That happens when resistors or other elements sit in parallel. The current splits among the branches in proportion to each branch conductance, not to reduce the total energy used by the circuit.

In a parallel circuit the supply current splits among the parallel paths. That split is the current divider rule, also called the current divider law.

In a parallel current divider, every branch shares the same two end nodes. That shared pair of nodes is what creates the parallel paths.

Branch currents can differ, but the voltage is the same across every path: $V_R_1 = V_R_2 = V_R_3$ …. etc. You do not need a separate voltage for each resistor. Branch currents then follow from KCL (Kirchhoff’s Current Law) and ohm’s law.

The equivalent resistance of a parallel group is always smaller than the smallest single branch resistance.

Current Divider Formula

A general current-divider formula is

$\begin{align*} I_X = I_T [\frac {R_T}{R_X}] \end{align*}$

Where,

$I_X$ = Current through any resistor in the parallel circuit = $\frac{V}{R_X}$

$I_T$ = Total current of the circuit = $\frac{V}{R_T}$

$R_T$ = Equivalent resistance of the parallel circuit

$V$ = Voltage across the parallel circuit = $I_T R_T$ = $I_X R_X$ (as the voltage is same across all the components of the parallel circuit)

In terms of impedance, the same rule is

$\begin{align*} I_X = I_T [\frac {Z_T}{Z_X}] \end{align*}$

In terms of admittance, it is

$\begin{align*} I_X = I_T [\frac {Y_X}{Y_T}] \,\,\,\, (as \,\, Z = \frac{1}{Y}) \end{align*}$

Current Divider Formula for RC Parallel Circuit

For the RC parallel circuit below, the resistor current is

$\begin{align*} I_R = I_T [\frac {Z_C}{R+Z_C}] \end{align*}$

where $Z_C$ is the impedance of the capacitor, $\frac{1}{j\omega C}$

which simplifies to

$\begin{align*} \begin{split*} & I_R = I_T [\frac {\frac{1}{j\omega C}}{R+\frac{1}{j\omega C}}]\\ = I_T [\frac {\frac{1}{j\omega C}}{\frac{j\omega CR+1}{j\omega C}}]\\ \end{split*} \end{align*}$

$\begin{align*} I_R = I_T [\frac{1}{1+j\omega RC}] \end{align*}$

Current Divider Rule Derivations

Take two resistors R₁and R₂ in parallel across a source of V volts.

Let the total current entering the pair be I_T. That current I_T splits into I₁and I₂, with I₁ in R₁and I₂ in R_2.

Kirchhoff’s current law then gives:

(1) $\begin{equation*} I_T = I_1+I_2 \end{equation*}$

(2) $\begin{equation*} I_1 = I_T-I_2 \end{equation*}$

(3) $\begin{equation*} I_2= I_T-I_1 \end{equation*}$

The equivalent resistance R_eq of the two parallel resistors is

$\begin{align*} R_e_q = R_1 // R_2 \end{align*}$

(4) $\begin{equation*} R_e_q = \frac {R_1 * R_2}{R_1 + R_2} \end{equation*}$

From Ohm’s law $I=\frac{V}{R}$ , the current in R₁ is

$\begin{align*} I_1 = \frac{V}{R_1} \end{align*}$

(5) $\begin{equation*} V = I_1 R_1 \end{equation*}$

The current in R₂ is

$\begin{align*} I_2 = \frac{V}{R_2} \end{align*}$

(6) $\begin{equation*} V = I_2 R_2 \end{equation*}$

Equations (5) and (6) give the same voltage, so

$\begin{align*} V = I_1 R_1 = I_2 R_2 \end{align*}$

$\begin{align*} I_1 = I_2 \frac{R_2}{R_1} \end{align*}$

Substitute that I₁into equation (1):

$\begin{align*} \begin{split*} & I_T = I_2\frac{R_2}{R_1}+I_2\\ = I_2 [\frac{R_2}{R_1}+1]\\ = I_2 [\frac{R_2+R_1}{R_1}] \end{split*} \end{align*}$

(7) $\begin{equation*} I_2 = I_T [\frac{R_1}{R_1+R_2}]\end{equation*}$

Then put I₂ into equation (2):

$\begin{align*} \begin{split*} & I_1 = I_T - I_T [\frac{R_1}{R_1+R_2}]\\ = I_T [1-\frac{R_1}{R_1+R_2}]\\ = I_T [\frac{R_1+R_2-R_1}{R_1+R_2}] \end{split*} \end{align*}$

(8) $\begin{equation*} I_1 = I_T [\frac{R_2}{R_1+R_2}] \end{equation*}$

Equations (7) and (8) say that, for two parallel resistors, each branch current equals the opposite-branch resistance over the sum of the two resistances, times the total current.

In general,

$\,\,Branch\,\,Current\,\,=\,\,Total\,\,Current*(\frac{resistance\,\,of\,\,opposite\,\,branch}{sum\,\,of\,\,the\,\,resistance\,\,of \,\,the\,\,two\,\,branch})$

Current Divider Examples

Current Divider for 2 Resistors in Parallel With Current Source

Example 1: 20 Ω and 40 Ω sit in parallel on a 20 A current source. Find each branch current.

Given data: R₁ = 20Ω, R₂ = 40Ω and I_T = 20 A

Current through resistor R₁ is given by

$\begin{align*} \begin{split} & I_1 = I_T [\frac{R_2}{R_1+R_2}] = 20[\frac{40}{20+40}] = 20[\frac{40}{60}] = 20[0.67] =13.33 A \end{split} \end{align*}$

(9) $\begin{equation*} I_1 = 13.33 A \end{equation*}$

Current through resistor R₂ is given by

$\begin{align*} \begin{split} & I_2 = I_T [\frac{R_1}{R_1+R_2}] = 20[\frac{20}{20+40}] = 20[\frac{20}{60}] = 20[0.33] =6.67 A \end{split} \end{align*}$

(10) $\begin{equation*} I_2 = 6.67 A \end{equation*}$

Adding equations (9) and (10) gives:

$\begin{align*} I_1 + I_2 = 13.33 + 6.67 = 20 A = I_T \end{align*}$

That matches Kirchhoff’s current law: the branch currents add to I_T. The 20 Ω branch takes the larger share because it is the smaller resistance.

Current Divider for 2 Resistors in Parallel With Voltage Source

Example 2: 10 Ω and 20 Ω sit in parallel on a 50 V voltage source. Find the total current and each branch current.

Given data: R₁ = 10Ω, R₂ = 20Ω and V = 50 V

The equivalent resistance of the parallel circuit is given by,

$\begin{align*} \begin{split} & R_e_q. = \frac{R_1*R_2}{R_1+R_2} = \frac{10*20}{10+20} = 6.67\Omega \end{split} \end{align*}$

According to ohm’s law, the total current flowing through the parallel circuit is given by,

$\begin{align*} \begin{split} & I_T = \frac{V}{R_e_q.} = \frac{50}{6.67} = 7.5 A \end{split} \end{align*}$

Now, according to current divider formula the Current through resistor R₁ is given by

$\begin{align*} \begin{split} & I_1 = I_T [\frac{R_2}{R_1+R_2}] = 7.5[\frac{20}{10+20}] = 7.5[\frac{20}{30}] = 7.5[0.67] = 5 A \end{split} \end{align*}$

(11) $\begin{equation*} I_1 = 5 A \end{equation*}$

Current through resistor R₂ is given by

$\begin{align*} \begin{split} & I_2 = I_T [\frac{R_1}{R_1+R_2}] = 7.5[\frac{10}{10+20}] = 7.5[\frac{10}{30}] = 7.5[0.33] = 2.5 A \end{split} \end{align*}$

(12) $\begin{equation*} I_2 = 2.5 A \end{equation*}$

Add equations (11) and (12):

$\begin{align*} I_1 + I_2 = 5 + 2.5 = 7.5 A = I_T \end{align*}$

The branch currents add to I_T, as Kirchhoff’s current law requires. The 10 Ω branch again takes the larger share.

Current Divider for 3 Resistors in Parallel

Example 3: 20 Ω, 30 Ω and 40 Ω sit in parallel on a 100 V supply. Find the total current and each branch current with the current-divider formula I_X = I_T (R_T / R_X).

Given data: R₁ = 20Ω, R₂ = 30Ω, R₃ = 40Ω and V = 100 V

The equivalent resistance of the parallel circuit is given by,

$\begin{align*} \begin{split} R_e_q. = \frac{R_1*R_2}{R_1+R_2} //R_3 = \frac{20*30}{20+30}//40 = \frac{600}{50}//40 & = 12//40\\ = \frac{12*40}{12+40} = \frac{480}{52} & = 9.231\Omega \\ R_e_q.= 9.231\Omega \end{split} \end{align*}$

According to ohm’s law, the total current flowing through the parallel circuit is given by,

$\begin{align*} \begin{split} & I_T = \frac{V}{R_e_q.} = \frac{100}{9.231} = 10.83 A \end{split} \end{align*}$

Now, according to current divider formula, the current through resistor R₁ is given by

$\begin{align*} \begin{split} & I_1 = I_T [\frac{R_e_q.}{R_1}] = 10.83[\frac{9.231}{20}] = 10.83[0.46] = 4.999 A \end{split} \end{align*}$

(13) $\begin{equation*} I_1 = 4.999 A \end{equation*}$

Current through resistor R₂ is given by

$\begin{align*} \begin{split} & I_2 = I_T [\frac{R_e_q.}{R_2}] = 10.83[\frac{9.231}{30}] = 10.83[0.31] = 3.33 A \end{split} \end{align*}$

(14) $\begin{equation*} I_1 = 3.33 A \end{equation*}$

Similarly, current through resistor R₃ is given by

$\begin{align*} \begin{split} & I_3 = I_T [\frac{R_e_q.}{R_3}] = 10.83[\frac{9.231}{40}] = 10.83[0.12] = 2.499 A \end{split} \end{align*}$

(15) $\begin{equation*} I_1 = 2.499 A \end{equation*}$

Add the three branch results:

$\begin{align*} I_1 + I_2 + I_3 = 4.999 + 3.33 + 2.499 = 10.83 A = I_T \end{align*}$

The three branch currents add to I_T.

In every example the smallest resistance takes the largest current. The source current equals the sum of the branch currents.

When You Can Use the Current Divider Rule

Use the current divider rule when:

Two or more elements share the same two nodes, whether the source is a voltage source or a current source.

The total current and the equivalent parallel resistance are known, so each branch current is I_T times R_eq / R_branch.

Two resistors are in parallel. Each branch current is a fraction of I_T. Equal resistors share the current equally.

Three or more resistors are in parallel. Use R_eq. in I_X = I_T (R_eq / R_X) as in example 3.