Electrical Engineering BMT & UFC › Circuit Analysis 2

Current Divider

A current divider is used to split a current in a parallel connection in a targeted way. Conversely, the current divider can be used to calculate how a current splits in a parallel connection. Like the voltage divider, the current divider is defined for the special case of two resistors. If the values of the two resistors change, the division of the current changes.

If two resistors R₁ and R₂ are connected in parallel, the following applies:

\[ \begin{gathered} U_1 = U_2 \\[4pt] I_1 = \frac{U_1}{R_1} \\[4pt] I_2 = \frac{U_2}{R_2} \\[4pt] I_0 = I_1 + I_2 \\[4pt] I_1 = \frac{U_1}{R_1} = \frac{U_2}{R_1} = \frac{R_2 \cdot I_2}{R_1} \\[4pt] I_0 = I_1 + I_2 = \frac{R_2}{R_1} I_2 + I_2 \\[4pt] \text{Expanded with } \frac{R_1}{R_1}\text{: } I_0 = \frac{R_2}{R_1} I_2 + \frac{R_1}{R_1} I_2 = \frac{R_1 + R_2}{R_1} I_2 \\[4pt] I_1 = I_0 \cdot \frac{R_2}{R_1 + R_2} \\[4pt] I_2 = I_0 \cdot \frac{R_1}{R_1 + R_2} \end{gathered} \]

We call the two bottom formulas the current divider formulas. They can be used to calculate a partial current without having to know any voltage in the network. We only need the current flowing into the parallel connection. In addition, we need the values of the two resistors.

These formulas only apply exactly to the circuit drawn. If the parallel connection is not in this form, it must first be brought into this form by combining resistors. If, for example, the partial current through R₁ in the following circuit is to be calculated with the current divider, R₂ and R₃ must first be combined into R₂₃ (series connection).

\[ \begin{gathered} R_{23} = R_2 + R_3 \\[6pt] I_1 = I_0 \cdot \frac{R_{23}}{R_1 + R_{23}} = I_0 \cdot \frac{R_2 + R_3}{R_1 + R_2 + R_3} \end{gathered} \]

In the current divider, the denominator always contains the sum of the resistances. The numerator contains the resistance through which the current does not flow – i.e. the other resistor.

Problem: Calculate the current I₂ in the following network:

Parallel connection of three equal resistors

\[ \begin{gathered} R_1 = R_2 = R_3 = 30\,\Omega \\[4pt] I_0 = 6\,\mathrm{A} \end{gathered} \]

Intuitive solution: the current divides equally among the three equal resistors. So 1/3 of the current I₀ flows in the middle branch. Thus I₂ = 2 A. For the computational solution, the resistors R₁ and R₃, which are connected in parallel, must first be combined. The following applies:

\[ R_{13} = R_1 \,||\, R_3 = \frac{R_1 \cdot R_3}{R_1 + R_3} = \frac{30\,\Omega \cdot 30\,\Omega}{30\,\Omega + 30\,\Omega} = 15\,\Omega \]

The simplified circuit now has the standard form for the current divider.

Current divider consisting of R2 and R13

\[ I_2 = I_0 \cdot \frac{R_{13}}{R_2 + R_{13}} = 6\,\mathrm{A} \cdot \frac{15\,\Omega}{15\,\Omega + 30\,\Omega} = 6\,\mathrm{A} \cdot \frac{1}{3} = 2\,\mathrm{A} \]

Simulation