The Voltage Divider - UPDATED 22 April 2020
See also https://www.electronics-tutorials.ws/dccircuits/voltage-divider.html.
Introduction
Any series network of resistances (or impedances, but that is a more advanced topic) can act as a voltage divider. The voltage divider comes from using Kirchhoff's Voltage Law on a series connection of resistances. Consider the diagram below. A voltage source \( V \) is driving a series network of resistors \( R_1, R_2, \ldots, R_n \), causing a current flow \( I \).
By Kirchhoff's Current Law, the same current must flow through all resistors.
By Kirchhoff's Voltage Law, the sum of the voltage drops across the resistors must equal the supply voltage.
In other words, \( V = V_1 + V_2 + \ldots + V_n \), where \( V_1 \) is the voltage drop across \( R_1 \), \( V_2 \) is the voltage drop across \( R_2 \), etc.
By Ohm's Law, the voltage \( V_m \) across any resistor \( R_m \) is given by \( V_m = I R_m \).
We can calculate the current \( I \) using Ohm's Law:
\( I = \frac{V}{R_{\mathrm{TOT}}} \),
where \( R_{\mathrm{TOT}} \) is the sum of all the series resistances i.e. \( R_{\mathrm{TOT}} = R_1 + R_2 + \ldots + R_n \).
Substituting \( I \) into \( V_m = I R_m \), we get:
\( V_m = V \cdot \frac{R_m}{R_{\mathrm{TOT}}} \)
Example: Two-Resistor Divider
A circuit with a voltage divider using two resistors is shown below.
In terms of the Introduction above, this is what we know:
| Quantity | Value |
|---|---|
| \( V \) | 100 V |
| \( R_1 \) | 75 Ω |
| \( R_2 \) | 60 Ω |
| \( R_{\mathrm{TOT}} \) | 135 Ω |
The table of calculations of the voltage drops across the resistors is shown below. Notice how the sum of the voltage drops across the resistors is the same as the supply voltage.
| Quantity | Value | Calculation |
|---|---|---|
| \( V_1 \) | 55.56 V | \( V_1 = V \cdot \frac{R_1}{R_{\mathrm{TOT}}} = 100 \cdot \frac{75}{135} = 100 \cdot 0.5556 \) |
| \( V_2 \) | 44.44 V | \( V_2 = V \cdot \frac{R_2}{R_{\mathrm{TOT}}} = 100 \cdot \frac{60}{135} = 100 \cdot 0.4444 \) |
| Total Voltage Drop | 100.0 V | \( V_1 + V_2 \) |
Our circuit is solved.
Example: Three-Resistor Divider
A circuit with a voltage divider using three resistors is shown below.
In terms of the Introduction above, this is what we know:
| Quantity | Value |
|---|---|
| \( V \) | 120 V |
| \( R_1 \) | 18 Ω |
| \( R_2 \) | 12 Ω |
| \( R_3 \) | 6 Ω |
| \( R_{\mathrm{TOT}} \) | 36 Ω |
The table of calculations of the voltage drops across the resistors is shown below. Notice how the sum of the voltage drops across the resistors is the same as the supply voltage.
| Quantity | Value | Calculation |
|---|---|---|
| \( V_1 \) | 60.00 V | \( V_1 = V \cdot \frac{R_1}{R_{\mathrm{TOT}}} = 120 \cdot \frac{18}{36} = 120 \cdot 0.5 \) |
| \( V_2 \) | 40.00 V | \( V_2 = V \cdot \frac{R_2}{R_{\mathrm{TOT}}} = 120 \cdot \frac{12}{36} = 120 \cdot 0.3333 \) |
| \( V_3 \) | 20.00 V | \( V_3 = V \cdot \frac{R_3}{R_{\mathrm{TOT}}} = 120 \cdot \frac{6}{36} = 120 \cdot 0.1667 \) |
| Total Voltage Drop | 120.00 V | \( V_1 + V_2 + V_3 \) |
The total voltage drop is the same as the supply voltage, which means our voltage drop calculations are correct.
Use in Circuit Analysis
Of course, "real" circuits are not just a single voltage source, however, if we know the total voltage across a series network, or total current through a series network, we can still use the tools about to solve for the voltages across each resistor.
Consider the circuit below. The voltage \( V \) represents the voltage across the series resistor network, and the current \( I \) represents the current through the series resistor network. The boxes have any arrangement of resistors, voltage sources, current sources, etc. But the key to this circuit is that if we know \( V \) or \( I \), we do not need to know how the unknown circuits are configured!
If we know \( V \), we can use the voltage divider equations to get the voltage across each resistor: \( V_m = V \cdot \frac{R_m}{R_{\mathrm{TOT}}} \).
If we know \( I \), we can use Ohm's Law to get the voltage across each resistor: \( V_m = I R_m \).
The rest of the circuit layout does not matter!!!