**The Voltage Divider Rule**

**What is the voltage divider rule? **

The voltage divider rule is a simple way of determining the output voltage across one of two impedances connected in series. It is a useful tool for circuit analysis and design.

**How is it used? **

The voltage divider rule can be used with resistive, inductive, or capacitive circuit elements. It can also be used with AC or DC input sources. The equation for calculating the output voltage is different, however, depending on the type of circuit element. Following are the three general cases of two like elements connected in series:

- Resistive divider:
The formula for determining the DC or AC output voltage of a resistive divider is:Vout = Vin*R2/(R1+R2)Example: In the following circuit, the output voltage would be: Vout = 9V*10K/(10K + 5K) = 6V

- Inductive divider:

Inductive dividers can be used with AC input signals. A DC input voltage would split according to the relative resistances of the two inductors by using the resistive divider formula above. The formula for determining the AC output voltage of an inductive divider (provided the inductors are separate, i.e. not wound on the same core, and have no mutual inductance) is:Vout = Vin*L2/(L1+L2)Example:In the following circuit, the output voltage would be: Vout = 10VAC*50mH/(50mH + 100mH) = 3.33VAC. Note that the output voltage is not dependent on the input frequency. However, if the reactance of the inductors is not high at the frequency of operation (i.e. inductance not large enough), there will be a very large current drawn by the shunt element.

- Capacitive divider:

Capacitive dividers can be used with AC input signals. A DC input voltage would not pass through the capacitors, so the DC case is not relevant. The formula for determining the AC output voltage of a capacitive divider is different from the resistive and inductive dividers, because the series element, C1 is in the numerator instead of the shunt element, as shown below:

Vout = Vin*C1/(C1+C2)Example: In the following circuit, the output voltage would be: Vout = 10VAC*0.022uF/(0.022uF + 0.01uF) = 6.875VAC. Note that the output voltage is not dependent on the input frequency. However, if the reactance of the capacitors is not small enough at the frequency of interest (i.e. if the capacitance not large enough), the output current capability will be very low.

Copyright © 1999, Randall Aiken. May not be reproduced in any form without written approval from Aiken Amplification.

*Revised 02/19/17*