Capacitors

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Three capacitors of equal capacitance are in a series circuit. If the total capacitance is , what is the capacitance of each individual capacitor?

$\frac{1}{36}F$

$\frac{1}{4}F$

Explanation

The formula for capacitors in a series is:

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}...$

Since we have three equal capacitors, we can say:

$\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}=3*\frac{1}{C}$

Using the value for total capacitance, we can find the value for the individual capacitance.

$\frac{1}{C_{eq}}=\frac{3}{C}$

$\frac{1}{12F}=\frac{3}{C}$

Three capacitors of equal capacitance are in a series circuit. If the total capacitance is , what is the capacitance of each individual capacitor?

$\frac{1}{36}F$

$\frac{1}{4}F$

Explanation

The formula for capacitors in a series is:

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}...$

Since we have three equal capacitors, we can say:

$\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}=3*\frac{1}{C}$

Using the value for total capacitance, we can find the value for the individual capacitance.

$\frac{1}{C_{eq}}=\frac{3}{C}$

$\frac{1}{12F}=\frac{3}{C}$

Which of these is the correct SI unit for capacitance?

Farads

Newtons

Ohms

Volts

Farad-Joules

Explanation

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.

Calculate the total capacitance of a circuit with the following three capacitors in parallel.

$C_1=10F,\ C_2=15F,\ C_3=5F$

Explanation

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

$C_{eq}=C_1+C_2+C_3$

$C_{eq}=10F+15F+5F=30F$

Calculate the total capacitance of a circuit with the following three capacitors in parallel.

$C_1=10F,\ C_2=15F,\ C_3=5F$

Explanation

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

$C_{eq}=C_1+C_2+C_3$

$C_{eq}=10F+15F+5F=30F$

Which of these is the correct SI unit for capacitance?

Farads

Newtons

Ohms

Volts

Farad-Joules

Explanation

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.

What is the total capacitance of a series circuit with capacitors of , , and ?

Explanation

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

Plug in our given values.

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}...$

$\frac{1}{C_{eq}}=\frac{1}{5F}+\frac{1}{1.1F}+\frac{1}{7.5F}$

$\frac{1}{C_{eq}}=1.24F$

$C_{eq}=0.8F$

What is the total capacitance of a series circuit with capacitors of , , and ?

Explanation

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

Plug in our given values.

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}...$

$\frac{1}{C_{eq}}=\frac{1}{5F}+\frac{1}{1.1F}+\frac{1}{7.5F}$

$\frac{1}{C_{eq}}=1.24F$

$C_{eq}=0.8F$

Three capacitors are in series. They have capacitances of , , and , respectively. What is their total capacitance?

Explanation

For capacitors in series the formula for total capacitance is:

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}...$

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.

$\frac{1}{C_{eq}}=\frac{1}{3F}+\frac{1}{2F}+\frac{1}{8F}$

$\frac{1}{C_{eq}}=0.958F$

$C_{eq}=1.04F$

Three capacitors are in series. They have capacitances of , , and , respectively. What is their total capacitance?

Explanation

For capacitors in series the formula for total capacitance is:

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}...$

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.

$\frac{1}{C_{eq}}=\frac{1}{3F}+\frac{1}{2F}+\frac{1}{8F}$

$\frac{1}{C_{eq}}=0.958F$

$C_{eq}=1.04F$

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