Potential Divider Circuits

A potential divider is two resistors in series across a supply voltage. The voltage is divided between them in proportion to their resistances.

The potential divider equation

$V_{\text{out}} = V_{\text{in}} \times \frac{R_2}{R_1 + R_2}$

where $V_{\text{out}}$ is the voltage across $R_2$.

Why it works

In a series circuit, the current through both resistors is the same:

$I = \frac{V_{\text{in}}}{R_1 + R_2}$

The voltage across $R_2$ is:

$V_2 = IR_2 = \frac{V_{\text{in}} R_2}{R_1 + R_2}$

The larger resistor gets the larger share of the voltage.

Worked example

$V_{\text{in}} = 12$ V, $R_1 = 4$ k$\Omega$, $R_2 = 8$ k$\Omega$.

$V_{\text{out}} = 12 \times \frac{8}{4 + 8} = 12 \times \frac{8}{12} = 8 \text{ V}$

Applications with sensors

Replace one resistor with a thermistor or LDR:

• As the sensor’s resistance changes, the voltage division changes.
• The output voltage varies with temperature or light level.
• This output can be used to trigger a relay, buzzer or electronic switch.

Common errors and how to correct them

• Using the wrong resistor in the formula. $V_{\text{out}}$ is the voltage across the resistor you take the output from.
• Forgetting that if a load is connected across $R_2$, the effective resistance changes and the simple formula no longer applies exactly.

How to apply this in an exam

Draw the circuit. Identify $R_1$ and $R_2$. Apply the formula. If one resistor is a sensor, explain how changing conditions change $V_{\text{out}}$.