The Potential Divider | CAIE 9702 Paper 1 Notes

Paper 1

Physics9702

The Potential Divider

Two series resistors split the input p.d. The larger resistance gets the larger share.

The output is always the p.d. across the resistor named as the output resistor.

Output is a fraction of the input

Two resistors share the input voltage in proportion to their resistances. The output wire simply takes R2's share.

The series current is $I=V_{\text{in}}/(R_1+R_2)$ . The output across $R_2$ is $IR_2$ , so:

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

Symbols

This expression assumes the output is unloaded, or connected to a device with resistance much larger than the divider resistors.

$V_{\text{out}}$ must lie between zero and $V_{\text{in}}$ .
Increasing the output resistance makes its fraction larger.
If the output is across the other resistor, that other resistor belongs in the numerator.

Your turn9702-style

A 3.0 kΩ resistor and a 1.0 kΩ resistor are in series across 12 V. Calculate the output p.d. across the 1.0 kΩ resistor. [2]

Show worked answer

$V_{\text{out}}=12\times1.0/(3.0+1.0)=3.0\ \text{V}$ .