When two 10 microfarad capacitors are connected in series, what is the net result?

When two 10 microfarad capacitors are connected in series, what is the net result?

• A. 0 mf
• B. 5 mf
• C. 10 mf
• D. 20 mf

Answer: (B)

When capacitors are connected in series, the total capacitance can be found using the formula: \[ \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} \] For two capacitors of equal capacitance, such as the two 10 microfarad capacitors in this case, the calculation becomes: \[ \frac{1}{C_{total}} = \frac{1}{10 \, \text{µF}} + \frac{1}{10 \, \text{µF}} = \frac{2}{10 \, \text{µF}} = \frac{1}{5 \, \text{µF}} \] To find the total capacitance, you would take the reciprocal of \(\frac{1}{5 \, \text{µF}}\): \[ C_{total} = 5 \, \text{µF} \] This formula and method show that when capacitors are in series, the resulting capacitance is less than the capacitance of the individual capacitors. Therefore, the correct answer of 5 microfarads reflects the proper outcome of this configuration. Red

When capacitors are connected in series, the total capacitance can be found using the formula:

[

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

]

For two capacitors of equal capacitance, such as the two 10 microfarad capacitors in this case, the calculation becomes:

[

\frac{1}{C_{total}} = \frac{1}{10 , \text{µF}} + \frac{1}{10 , \text{µF}} = \frac{2}{10 , \text{µF}} = \frac{1}{5 , \text{µF}}

]

To find the total capacitance, you would take the reciprocal of (\frac{1}{5 , \text{µF}}):

[

C_{total} = 5 , \text{µF}

]

This formula and method show that when capacitors are in series, the resulting capacitance is less than the capacitance of the individual capacitors. Therefore, the correct answer of 5 microfarads reflects the proper outcome of this configuration. Red