Capacitive Voltage Divider

Consider the two capacitors, C1 and C2 connected in series across an alternating supply of 10 volts. As the two capacitors are in series, the charge Q on them is the same, but the voltage across them will be different and related to their capacitance values, as V = Q/C.

Voltage divider circuits may be constructed from reactive components just as easily as they may be constructed from resistors as they both follow the voltage divider rule. Take this capacitive voltage divider circuit, for instance.

The voltage across each capacitor can be calculated in a number of ways. One such way is to find the capacitive reactance value of each capacitor, the total circuit impedance, the circuit current and then use them to calculate the voltage drop.

Capacitive Reactance Formula

• Where:
•    Xc = Capacitive Reactance in Ohms, (Ω)
•    π (pi) = a numeric constant of 3.142 (or 22÷7)
•    ƒ = Frequency in Hertz, (Hz)
•    C = Capacitance in Farads, (F)

Capacitive Voltage Divider Example No1

Using the two capacitors of 10uF and 22uF in the series circuit above, calculate the rms voltage drops across each capacitor when subjected to a sinusoidal voltage of 10 volts rms at 80Hz.

 Capacitive Reactance of 10uF capacitor

 Capacitive Reactance of 22uF capacitor

 Total capacitive reactance of series circuit – Note that reactance’s in series are added together just like resistors in series.

or:

 Circuit current

 Then the voltage drop across each capacitor in series capacitive voltage divider will be:

When the capacitor values are different, the smaller value capacitor will charge itself to a higher voltage than the larger value capacitor, and in our example above this was 6.9 and 3.1 volts respectively. Since Kirchoff’s voltage law applies to this and every series connected circuit, the total sum of the individual voltage drops will be equal in value to the supply voltage, VS and 6.9 + 3.1 does indeed equal 10 volts.