Looking for an all-in-one current divider calculator to estimate the current flowing through each branch of a resistive, inductive, or capacitive circuit? You're in the right place. Let's quickly inform you what our tool can do:

  • It's a current branch calculator.
  • Explains the current divider rule and how current divides in a resistive circuit with an example.
  • Guides you on how to derive the current divider formula for resistive circuits, inductive circuits, and capacitive circuits.
  • Helps you understand how other components influence the current flowing through a component in a parallel circuit.

In a nutshell, our tool is a comprehensive current division calculator!

What is a current divider?

Any circuit that divides current into various paths is a current divider circuit. The magnitude of current passing through a particular path depends on the impedance of that path. Unlike the voltage divider rule, current and impedance have an inverse relationship – the more the impedance of a path, the lesser the current passing through it.

For example, in a parallel connection of two resistors of equal resistance, the current from the source divides equally and flows through each resistor.

💡 A current divider circuit is also known as a parallel circuit.

Let's take a look at how current divides in a resistive circuit, inductive circuit, and capacitive circuit.

What is the current divider rule?

When we connect two components providing parallel resistance (or impedance in AC circuits), the current in any branch is a fraction of the total current. For example, in a 1-ampere DC parallel circuit with a 1Ω-resistor in each of the two branches, the current flowing through the branches is 0.5 A.

How does current divide in a parallel resistive circuit?

According to Ohm's law, in a parallel resistive circuit, the current flowing through a path is:

Ipath=VRpathI_\text{path} = \frac{V}{R_\text{path}}

where VV is the voltage across the path, and RpathR_\text{path} is the effective resistance of the path.

Example of a parallel resistive circuit with upto n resistors.
Example of a parallel resistive circuit with upto n resistors. (Source: wikimedia.org.)

In any parallel circuit, the voltage remains the same across two branches, and it is equal to the product of the source current II and the circuit's effective resistance ReqR_\text{eq}:

V=IReqV = IR_\text{eq}

Consider a parallel resistive circuit of two resistors, R1R_\text{1} and R2R_\text{2}. The effective resistance, ReqR_\text{eq} of the given parallel circuit is:

1Req=1R1+1R2\frac{1}{R_\text{eq}} = \frac{1}{R_1}+\frac{1}{R_2}

On rearranging:

Req=R1R2R1+R2R_\text{eq} = \frac{R_1 R_2}{ R_1 + R_2}

Thus, the voltage across the paths is:

V=IR1R2R1+R2V = I \frac{R_1 R_2}{ R_1 + R_2}

Now, divide the voltage across the first path by its effective resistance to obtain the current divider formula for R1R_\text{1}:

I1=IR1R1R2R1+R2I_1 =\frac{ I}{R_1} \frac{R_1 R_2}{ R_1 + R_2}

On simplifying:

I1=IR2R1+R2I_1 = I \frac{ R_2}{ R_1 + R_2}

Similarly, the current divider equation for the resistor R2R_\text{2} is:

I2=IR1R1+R2I_2 = I \frac{ R_1}{ R_1 + R_2}

✅ Hence, the current passing through a resistor is the product of total current and the ratio of the product of other resistances to the effective resistance.

If you want to add more resistors, you can repeat the process, although the formulas may grow big. But don't worry – simply use our current divider calculator, and you'll get the result in a second!

What is the current divider formula for an inductive circuit?

Example of a parallel inductive circuit with upto n inductors
Example of a parallel inductive circuit with upto n inductors. (Source: wikimedia.org.)

For a parallel inductive circuit, we can consider the current division only in an alternating current (AC). By analogy to the resistive current divider, we can apply Ohm's law to find the average current flowing through an inductance LL using the inductive reactance XLX_\text{L}:

I=VXL=V2πfLI = \frac{V}{X_\text{L}} = \frac{V}{2\pi fL}

where ff is the AC signal frequency.

Consider a parallel inductive circuit consisting of two inductors, L1L_\text{1} and L2L_\text{2}. We calculate the effective inductance LeqL_\text{eq} as follows:

1Leq=1L1+1L2\frac{1}{L_\text{eq}} = \frac{1}{L_1}+\frac{1}{L_2}

On rearranging:

Leq=L1L2L1+L2L_\text{eq} = \frac{L_1 L_2}{ L_1 + L_2}

Thus, the voltage across the path is:

V=2πfIL1L2L1+L2V = 2\pi fI \frac{L_1 L_2}{ L_1 + L_2}

Now, divide the voltage across the first path by its reactance to obtain the current flowing through L1L_\text{1}:

I1=2πfI2πfL1L1L2L1+L2I_1 = \frac{2\pi fI}{2\pi fL_1} \frac{L_1 L_2}{ L_1 + L_2}

Or:

I1=IL2L1+L2I_1 = I \frac{ L_2}{ L_1 + L_2}

Similarly, the average current passing through the inductor L2L_\text{2} is:

I2=IL1L1+L2I_2 = I \frac{ L_1}{ L_1 + L_2}

💡 The current that passes through a path consisting of an inductor is the product of the total current and the ratio of the product of inductances to the effective inductance.

What is the current divider equation for a capacitive circuit?

Similar to the inductive circuit, the current division in a capacitive one is possible with an AC signal. We can use the capacitive reactance, XCX_\text{C}, to express the average current passing through a loop with capacitance CC as:

I=VXC=V12πfC=2πfCV.I = \frac{V}{X_\text{C}} = \frac{V}{\frac{1}{2\pi fC}} = 2\pi fCV.

The effective capacitance of a parallel connection is the sum of the individual capacitances:

Ceq=C1+C2C_\text{eq} = C_1 + C_2

The voltage across the first branch of the parallel capacitive circuit is:

V=I2πf(C1+C2)V = \frac {I} { 2\pi f (C_1 + C_2)}

Now, divide the voltage across the first path by its reactance to obtain the current divider equation for C1C_\text{1}:

I1=VXC1=I2πf(C1+C2)12πfC1=IC1C1+C2\begin{aligned} I_1 &= \frac{V}{X_\text{C1}} = \frac{\frac {I} { 2\pi f (C_1 + C_2)}}{\frac{1}{2\pi fC_1}} \\[1em] &= I\frac{C_1}{ C_1 + C_2} \end{aligned}

And the current passing through C2C_\text{2} is:

I2=IC2C1+C2I_2 = I\frac{C_2}{ C_1 + C_2}

✅ Hence, the current passing through a path in a parallel capacitive circuit is the product of total current and the path's capacitance divided by the circuit's effective capacitance.

That's all you need to know about formulas. Now let's try to solve some computational examples and learn how to use Omni's current divider calculator.

Instructions to use the current divider calculator

Imagine we have a four resistor setup (R₁ = 20 Ω, R₂ = 40 Ω, R₃ = 80 Ω, R₄ = 100 Ω), and we need to know how the current divides into each branch when we arrange them in parallel. Let's say the source current (I) equals 1 A. Here is how our current division calculator works:

  1. Select the Resistive circuit type. It's a default option.
  2. Enter the current value (1 A).
  3. Input the consecutive resistances into the corresponding resistors' fields. The new ones will appear as you type the previous ones.
  4. That's all! The current divider calculator does the rest of the job. As a result, you can see a table with your resistors and the approximate values of corresponding currents flowing through each path:
    • I₁ = 0.513 A;
    • I₂ = 0.256 A;
    • I₃ = 0.128 A; and
    • I₄ = 0.103 A.

FAQ

How do I calculate the current passing through a branch in a resistive circuit?

To calculate the current through a branch of any circuit:

  1. Work out the effective resistance (Req) of the whole circuit.
  2. Estimate the voltage (V) across the branch using Ohm's law: V = IReq, where I is the source current.
  3. Divide voltage by the branch's resistance to obtain the current passing through it: I = V/Rbranch if it is a resistive circuit.

Why do we multiply the voltage by capacitance to find the branch current in capacitive circuits?

The charge stored in a capacitor is the product of the capacitance, and the voltage applied: Q = CV. The electric current is the rate of electric charge: I = Q/t. That's why we multiply the capacitance with the applied voltage to estimate the branch current.

Is a series circuit a current divider?

No. The current flows through all components and is the same across all elements in a series circuit. The series circuit is a voltage divider.

Can I use the current divider rule for series circuits?

You don't need to use the current divider rule for series circuits. In a series circuit, voltage divides across various components, and the same current flows through all of them.

Mehjabin Abdurrazaque and Wojciech Sas, PhD candidate
Circuit type
Resistive
Current (I)
A
Resistor 1 (R₁)
Ω
Resistor 2 (R₂)
Ω
You can add up to 10 resistors, fields will appear as you need them.
Input at least one resistor to obtain a result.
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