If a circuit has two capacitors in series with a total capacitance (CT) of 16.67 µF and one capacitor (C1) of 25 µF, what is the value of the second capacitor (C2)?

To determine the value of the second capacitor in a series circuit, it's essential to utilize the formula for total capacitance in series, which is given by:

[

\frac{1}{CT} = \frac{1}{C1} + \frac{1}{C2}

]

In this scenario, the total capacitance (CT) is 16.67 µF, and one of the capacitors (C1) is 25 µF.

Start by substituting the known values into the formula:

[

\frac{1}{16.67} = \frac{1}{25} + \frac{1}{C2}

]

To isolate ( \frac{1}{C2} ), rearrange the equation:

[

\frac{1}{C2} = \frac{1}{16.67} - \frac{1}{25}

]

Now calculate ( \frac{1}{16.67} ) and ( \frac{1}{25} ):

• ( \frac{1}{16.67} \approx 0.060 )

• ( \frac{1}{25} = 0.04 )

Now, performing the subtraction: