The correct unit for capacitance is the Farad, named after the English physicist Michael Faraday.

Newtons are used to measure force. Ohms measure resistance and volts measure electrical energy or potential. Joules are a unit of energy, but Farad-Joules are not a recognizable unit.

Dielectrics are insulating materials that can be polarized, and can be used to increase the separability of the two plates of a capacitor. When a dielectric is absent, air is present between the capacitor plates. At high voltages, when the electric field between the plates is incredibly strong, air begins to break down and becomes unable to prevent the flow of electricity between the capacitor plates. When this occurs, the capacitor cannot store energy, nor can it properly function. The presence of a dielectric eliminates this problem, allowing the capacitor to store much more energy.

For capacitors in series the formula for total capacitance is:

\(\displaystyle \frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}...\)

Note that this formula is similar to the formula for total resistance in parallel. Using the values for each individual capacitor, we can solve for the total capacitance.

\(\displaystyle \frac{1}{C_{eq}}=\frac{1}{3F}+\frac{1}{2F}+\frac{1}{8F}\)

\(\displaystyle \frac{1}{C_{eq}}=0.958F\)

\(\displaystyle C_{eq}=1.04F\)

For capacitors in parallel the formula for total capacitance is:

\(\displaystyle C_{eq}=C_1+C_2+C_3...\)

Note that this formula is similar to the formula for total resistance in series. Using the values for each individual capacitor, we can solve for the total capacitance.

\(\displaystyle C_{eq}=3F+2F+8F\)

\(\displaystyle C_{eq}=13F\)

The formula for capacitors in parallel is:

\(\displaystyle C_{eq}=C_1+C_2+C_3...\)

Our three capacitors all have equal capacitance values. We can simply add them together to find the total capacitance.

\(\displaystyle C_{eq}=4F+4F+4F\)

\(\displaystyle C_{eq}=12F\)

The total capacitance of a series circuit is \(\displaystyle \frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}...\)

Plug in our given values.

\(\displaystyle \frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}...\)

\(\displaystyle \frac{1}{C_{eq}}=\frac{1}{5F}+\frac{1}{1.1F}+\frac{1}{7.5F}\)

\(\displaystyle \frac{1}{C_{eq}}=1.24F\)

\(\displaystyle C_{eq}=0.8F\)

To calculate the total capacitance for capacitors in parallel, simply sum the value of each individual capacitor.

\(\displaystyle C_{eq}=C_1+C_2+C_3\)

\(\displaystyle C_{eq}=10F+15F+5F=30F\)

To find the total capacitance for capacitors in series, we must sum the inverse of each individual capacitance and take the reciprocal of the result.

\(\displaystyle \frac{1}{C_{eq}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}\)

\(\displaystyle \frac{1}{C_{eq}}=\frac{1}{10F}+\frac{1}{15F}+\frac{1}{5F}\)

\(\displaystyle \frac{1}{C_{eq}}=\frac{11}{30}=0.367\)

Remember, you must still take the final reciprocal!

\(\displaystyle C_{eq}=\frac{30}{11}=2.73F\)

The formula for capacitors in parallel is:

\(\displaystyle C_{eq}=C_1+C_2+C_3...\)

Plug in our given values:

\(\displaystyle C_{eq}=2F+12F+8F\)

\(\displaystyle C_{eq}=22F\)

The formula for capacitors in series is:

\(\displaystyle \frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}...\)

Plug in our given values:

\(\displaystyle \frac{1}{C_{eq}}=\frac{1}{2F}+\frac{1}{12F}+\frac{1}{8F}\)

\(\displaystyle \frac{1}{C_{eq}}=0.70833\frac{1}{F}\)

\(\displaystyle C_{eq}=1.41F\)