<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Skip Navigation
Our Terms of Use (click here to view) have changed. By continuing to use this site, you are agreeing to our new Terms of Use.

Capacitors Circuits

Circuits containing capacitors are easy to understand, take a closer look at them!

Atoms Practice
Estimated5 minsto complete
Practice Capacitors Circuits
This indicates how strong in your memory this concept is
Estimated5 minsto complete
Practice Now
Turn In
Capacitors Circuits

When a capacitor is placed in a circuit, current does not actually travel across it. Rather, equal and opposite charge begins to build up on opposite sides of the capacitor --- mimicking a current --- until the electric field in the capacitor creates a potential difference across it that balances the voltage drop across any parallel resistors or the voltage source itself (if there are no resistors in parallel with the capacitor). The ratio of charge on a capacitor to potential difference across it is called capacitance.

It is important to break down a complicated circuit into the equivalent capacitance using the rules for capacitors in series and capacitors in parallel. Also remember that capacitors in parallel have the same voltage while capacitors in series have the same charge.

Key Equations
\begin{align*} C = \frac{Q}{V} && \text{Definition of Capacitance} \end{align*}

\begin{align*} C_{\text{parallel}} &= C_1 + C_2 + C_3 + \dots && \text{Capacitors in parallel add like resistors in series}\\ \frac{1}{C_{\text{series}}} &= \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}+ \dots && \text{Capacitors in series add like resistors in parallel} \end{align*}


In the circuit shown below, determine (a) the total capacitance and (b) the total charge stored.

(a): In solving this problem, we'll call the \begin{align*}20\;\mu\text{F}\end{align*} capacitor \begin{align*}C_1\end{align*}, the \begin{align*}60\;\mu\text{F}\end{align*} capacitor \begin{align*}C_2\end{align*}, the \begin{align*}40\;\mu\text{F}\end{align*} capacitor \begin{align*}C_3\end{align*}, and the \begin{align*}100\;\mu\text{F}\end{align*} capacitor \begin{align*}C_4\end{align*}.

Our first step will be to find the equivalent capacitance of \begin{align*}C_2\end{align*} and \begin{align*}C_3\end{align*}.

\begin{align*} \frac{1}{C_{2,3}}&=\frac{1}{C_2}+\frac{1}{C_3}\\ \frac{1}{C_{2,3}}&=\frac{1}{60\;\mu\text{F}} + \frac{1}{40\;\mu\text{F}}\\ \frac{1}{C_{2,3}}&=\frac{5}{120\;\mu\text{F}}\\ C_{2,3}&=24\;\text{F}\\ \end{align*}

Next, we'll combine the capacitance of \begin{align*}C_{2,3}\end{align*} and \begin{align*}C_4\end{align*}.

\begin{align*} C_{2,3,4}&=C_{2,3} + C_4\\ C_{2,3,4}&=24\;\mu\text{F} + 100\;\mu\text{F}\\ C_{2,3,4}&=124\;\mu\text{F}\\ \end{align*}

Finally, we can combine \begin{align*}C_{2,3,4}\end{align*} with \begin{align*}C_1\end{align*} to find the total capacitance.

\begin{align*} \frac{1}{C_{eq}}&=\frac{1}{C_1} + \frac{1}{C_{2,3,4}}\\ \frac{1}{C_{eq}}&=\frac{1}{20\;\mu\text{F}} + \frac{1}{124\;\mu\text{F}}\\ \frac{1}{C_{eq}}&=.058\;\mu\text{F}^{-1}\\ C_{eq}&=17.22\;\mu\text{F}\\ \end{align*}

(b): Now we can use this value to find the total charge stored on all the capacitors by also using the voltage provided on the diagram.

\begin{align*} Q&=CV\\ Q&=17.22\;\mu\text{F} * 10\;\text{V}\\ Q&=172.2\mu\text{C} \end{align*}




Circuit Constructoin Kit (AC+DC) (PhET Simlation)


  1. Consider the figure above with switch, \begin{align*}S\end{align*}, initially open and the power supply set to 24 V:
    1. What is the voltage drop across the \begin{align*}20 \Omega \end{align*} resistor?
    2. What current flows thru the \begin{align*}60 \Omega\end{align*} resistor?
    3. What is the voltage drop across the \begin{align*}20\end{align*} microfarad capacitor?
    4. What is the charge on the capacitor?
    5. How much energy is stored in that capacitor?
    6. Find the capacitance of capacitors \begin{align*}B\end{align*}, \begin{align*}C\end{align*}, and \begin{align*}D\end{align*} if compared to the \begin{align*}20 \mu \mathrm{F}\end{align*}capacitor where...
      1. \begin{align*}B\end{align*} has twice the plate area and half the plate separation
      2. \begin{align*}C\end{align*} has twice the plate area and the same plate separation
      3. \begin{align*}D\end{align*} has three times the plate area and half the plate separation
  2. Now the switch in the previous problem is closed.
    1. What is the total capacitance of branch with B and C?
    2. What is the total capacitance of the circuit?
    3. What is the voltage drop across capacitor \begin{align*}B\end{align*}?

Review (Answers)

  1. a. \begin{align*}6\mathrm{V}\end{align*} b. \begin{align*}0.3\mathrm{A}\end{align*} c. \begin{align*}18\mathrm{V}\end{align*} d. \begin{align*}3.6 \times 10^{-4}\mathrm{C}\end{align*} e. \begin{align*}3.2 \times 10^{-3}\mathrm{J}\end{align*} f. i) \begin{align*}80 \mu \mathrm{F}\end{align*} ii) \begin{align*}40\mu \mathrm{F}\end{align*} iii) \begin{align*}120 \mu \mathrm{F}\end{align*}
  2. a. \begin{align*}26.7\mu \mathrm{F}\end{align*} b. \begin{align*}166.7\mu \mathrm{F}\end{align*}

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Please to create your own Highlights / Notes
Show More

Image Attributions

Explore More

Sign in to explore more, including practice questions and solutions for Capacitors Circuits.
Please wait...
Please wait...