What is the total capacitance when three capacitors valued at 100μF, 100μF, and 50μF are connected in series?

What is the total capacitance when three capacitors valued at 100μF, 100μF, and 50μF are connected in series?

• A. 10 μF
• B. 25 μF
• C. 50 μF
• D. 75 μF

Answer: (B)

To determine the total capacitance of capacitors connected in series, the following formula is applied: \[ \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \] where \( C_1, C_2, \) and \( C_3 \) are the capacitances of the individual capacitors. In this scenario, the values provided are 100 μF, 100 μF, and 50 μF. Substituting these values into the formula gives: \[ \frac{1}{C_{total}} = \frac{1}{100} + \frac{1}{100} + \frac{1}{50} \] This can be rewritten with a common denominator: \[ \frac{1}{C_{total}} = \frac{1}{100} + \frac{1}{100} + \frac{2}{100} = \frac{1 + 1 + 2}{100} = \frac{4}{100} \] Now, simplifying this results in: \[ \frac{1}{C_{total}} = \

To determine the total capacitance of capacitors connected in series, the following formula is applied:

[

\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

]

where ( C_1, C_2, ) and ( C_3 ) are the capacitances of the individual capacitors. In this scenario, the values provided are 100 μF, 100 μF, and 50 μF.

Substituting these values into the formula gives:

[

\frac{1}{C_{total}} = \frac{1}{100} + \frac{1}{100} + \frac{1}{50}

]

This can be rewritten with a common denominator:

[

\frac{1}{C_{total}} = \frac{1}{100} + \frac{1}{100} + \frac{2}{100} = \frac{1 + 1 + 2}{100} = \frac{4}{100}

]

Now, simplifying this results in:

[

\frac{1}{C_{total}} = \