The Current Divider - NEW 22 April 2020
See also https://www.electronics-tutorials.ws/dccircuits/current-divider.html.
Introduction
Any parallel network of resistances (or impedances, but that is a more advanced topic) can act as a current divider. The current divider comes from using Kirchhoff's Current Law on a parallel connection of resistances. Consider the diagram below. A current source \( I \) is driving a parallel network of resistors \( R_1, R_2, \ldots, R_n \), causing a voltage drop \( V \).
By Kirchhoff's Voltage Law, the same voltage must be across all resistors.
By Kirchhoff's Current Law, the sum of the currents through the resistors must equal the supply current.
In other words, \( I = I_1 + I_2 + \ldots + I_n \), where \( I_1 \) is the current through \( R_1 \), \( I_2 \) is the current through \( I_2 \), etc.
By Ohm's Law, the current \( I_m \) across any resistor \( R_m \) is given by \( I_m = \frac{V}{R_m} \).
We can calculate the voltage \( V \) using Ohm's Law:
\( V = I R_{\mathrm{TOT}} \),
where \( R_{\mathrm{TOT}} \) is the parallel sum of all the parallel resistances i.e. \( R_{\mathrm{TOT}} = (R_1^{-1} + R_2^{-1} + \ldots + R_n^{-1})^{-1} \).
Substituting \( V \) into \( I_m = \frac{V}{R_m} \), we get:
\( I_m = I \cdot \frac{R_{\mathrm{TOT}}}{R_m} \)
Example: Two-Resistor Divider
A circuit with a current divider using two resistors is shown below.
In terms of the Introduction above, this is what we know:
| Quantity | Value |
|---|---|
| \( I \) | 6 A |
| \( R_1 \) | 10 Ω |
| \( R_2 \) | 12 Ω |
| \( R_{\mathrm{TOT}} \) | 5.455 Ω |
| Parallel Voltage Drop | 32.73 V |
The table of calculations of the currents through the resistors is shown below. Notice how the sum of the currents through the the resistors is the same as the supply current.
| Quantity | Value | Calculation |
|---|---|---|
| \( I_1 \) | 3.273 A | \( V_1 = V \cdot \frac{R_{\mathrm{TOT}}}{R_1} = 6 \cdot \frac{5.455}{10} = 6 \cdot 0.5455 \) |
| \( I_2 \) | 2.728 A | \( V_2 = V \cdot \frac{R_{\mathrm{TOT}}}{R_2} = 6 \cdot \frac{5.455}{12} = 6 \cdot 0.4546 \) |
| Total Current | 6.001 A | \( I_1 + I_2 \) |
| Parallel Voltage Drop | 32.74 V | Multiply Total Current by \( R_{\mathrm{TOT}} \) |
While the currents and parallel voltage drop don't quite balance, the difference is due to round-off error. In any case, our circuit is solved.
The circuit with all solved voltages and currents is shown below.
Example: Three-Resistor Divider
A circuit with a current divider using three resistors is shown below.
In terms of the Introduction above, this is what we know:
| Quantity | Value |
|---|---|
| \( I \) | 2.5 A |
| \( R_1 \) | 40 Ω |
| \( R_2 \) | 8 Ω |
| \( R_3 \) | 20 Ω |
| \( R_{\mathrm{TOT}} \) | 5 Ω |
| Parallel Voltage Drop | 12.5 V |
The table of calculations of the currents through the resistors is shown below. Notice how the sum of the currents through the the resistors is the same as the supply current.
| Quantity | Value | Calculation |
|---|---|---|
| \( I_1 \) | 0.3125 A | \( I_1 = I \cdot \frac{R_{\mathrm{TOT}}}{R_1} = 2.5 \cdot \frac{5}{40} = 2.5 \cdot 0.125 \) |
| \( I_2 \) | 1.5625 A | \( I_2 = I \cdot \frac{R_{\mathrm{TOT}}}{R_2} = 2.5 \cdot \frac{5}{8} = 2.5 \cdot 0.625 \) |
| \( I_3 \) | 0.6250 A | \( I_3 = I \cdot \frac{R_{\mathrm{TOT}}}{R_3} = 2.5 \cdot \frac{5}{20} = 2.5 \cdot 0.25 \) |
| Total Current | 2.5000 A | \( I_1 + I_2 + I_3 \) |
| Parallel Voltage Drop | 12.5 V | Multiply Total Current by \( R_{\mathrm{TOT}} \) |
In this case, the currents and voltage drop balanced perfectly. Our circuit is solved.
The circuit with all solved voltages and currents is shown below.
