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)?

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)?

• A. 10 µF
• B. 50 µF
• C. 25 µF
• D. 16.67 µF

Answer: (B)

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:

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: