Energy in Capacitors

Section 34.13 Energy in Capacitors

A capacitor is an energy-storing device. By storing charges separated by a distance, the capacitor essentially stores energy in the potential energy of the charges, or equivalently in the electric field of the space between plates.

🔗

One way to easily figure out the energy stored in a capacitor is to use energy conservation in the discharging circuit. Connect a charged capacitor to a resistor \(R\) and let current flow in the simple RC-circuit and determine the net energy dissipated in the resistor. When current \(I(t)\) passes through a resistor, the instantaneous power-dissipation rate, \(P(t) = I(t)^2 R\text{,}\) as we have seen before.

🔗

We have already found the following expression for current at instant \(t\) in discharging circuit in which the initial charg eon the capacitor is \(Q_0\text{.}\)

\begin{equation*} I(t) = \dfrac{Q_0}{RC}\, \exp{\left( -t/RC\right ) }. \end{equation*}

Therefore, instantaneous power dissipated in the resistor will be

\begin{equation} P(t) = \dfrac{Q_0^2}{RC^2}\, \exp{\left( -2t/RC\right ) }.\tag{34.57} \end{equation}

🔗

This equation tells us the rate at which energy is being discharged by the capacitor. Therefore, energy discharged between \(t\) and \(t+\Delta t\) will be

\begin{equation*} \Delta U = P(t)\Delta t. \end{equation*}

Therefore, integrating Eq. (34.57) from \(t=0\) to \(t=T\) will give us the amount of energy discharged during this time.

\begin{equation} U = \int_0^T P(t)\; dt.\tag{34.58} \end{equation}

You can show that this leads to

\begin{equation} U = U_0\left( 1 - e^{-2T/RC}\right),\tag{34.59} \end{equation}

where \(U_0\) is the energy in the capacitor at time \(t=0\text{,}\) as we will interpret below.

\begin{equation*} U_0 = \frac{1}{2}Q_0 V_0. \end{equation*}

🔗

By taking \(T=\infty\) we get all energy that was dissipated in the resistor.

\begin{equation*} E_\text{dissipated} = \dfrac{1}{2}\dfrac{Q_0^2}{C}. \end{equation*}

But, by conservation of energy in the circuit, this energy must have been originally stored in the capacitor. Therefore, a capacitor of capacitance \(C\) charged to \(Q_0\) stores the following energy. Since this energy is potential energy, we use symbol \(U\) for it.

\begin{equation*} U_\text{in capacitor} = \dfrac{1}{2}\dfrac{Q_0^2}{C}. \end{equation*}

By using the capacitor formula, \(Q=CV\text{,}\) we can write this in other forms.

\begin{equation} U_\text{in capacitor} = \dfrac{1}{2}\dfrac{Q_0^2}{C} = \dfrac{1}{2}Q_0V_0 = \dfrac{1}{2}CV_0^2.\tag{34.60} \end{equation}

🔗

Example 34.80. Energy from Discharging a Capacitor.

(a) A capacitor of capacitance \(2.0\text{ mF}\) is charged by a \(1.5\text{-V}\) battery. How much total energy will be released when this capacitor is discharged?

🔗

(b) If discharging circuit has resistance \(5.0\;\Omega\text{,}\) how much time will it take to get \(70\%\) of the energy?

🔗

Answer.

(a) \(2.25\text{ mJ}\text{,}\) (b) \(1.78\text{ ms}\text{.}\)

🔗

Solution 1. (a)

The energy released will be equal to the energy stored in the capacitor. Since \(C\) and \(V_0\) are given, we use the formula using these.

\begin{equation*} U_0 = \frac{1}{2}CV_0^2 = \frac{1}{2}\times 2.0\text{ mF}\times (1.5\text{ V})^2 = 2.25\text{ mJ}. \end{equation*}

🔗

Solution 2. (b)

Energy released in time \(t=0\) to \(t=T\) is

\begin{equation*} U = U_0 \left( 1 - e^{-2t/RC}\right). \end{equation*}

We are given the following.

\begin{equation*} \frac{U}{U_0} = 0.3,\ \ RC = 5\times 2\times 10^{-3} = 0.01\text{ s}. \end{equation*}

Therefore, we have the following equation to solve.

\begin{equation*} 0.3 = 1 - e^{-200\,t}. \end{equation*}

We first isolate the exponential on one side and take natural log to solve this.

\begin{equation*} t = -\frac{ \ln\;0.7 }{200} = 1.78\text{ ms}. \end{equation*}

🔗

Prev Top Next