# Total Energy Stored - Circuit with Capacitors and Inductors

Find the total energy stored in the circuit.

Fig. (1-28-1) - The circuit

Solution
The circuit contains only dc sources. Recall that an inductor is a short circuit to dc and a capacitor is an open circuit to dc. These can be easily verified from their current-voltage characteristics. For an inductor, we have $v(t)=L \frac{d i(t)}{dt}$ . Since a dc current does not vary with time, $\frac{d i(t)}{dt}=0$ . Hence, the voltage across the inductor is zero for any dc current. This is to say that dc current passes through the inductor without any voltage drop, exactly similar to a short circuit. For a capacitor, the current-voltage terminal characteristics is $i(t)=L \frac{d v(t)}{dt}$ . Voltage drop across passive elements due to dc currents does not vary with time. Therefore, $\frac{d v(t)}{dt}=0$ and consequently the current of the capacitor is zero. This is to say that dc current does not pass through the capacitor regardless of the voltage amount. This is similar to the behavior of an open circuit. Please note that unlike dc current, ac current passes through capacitors in general.

To find dc current of the inductors and dc voltage drop across the capacitors, we replace them with their equivalent elements, short circuits and open circuits, respectively, as shown in Fig. (1-28-2).

Fig. (1-28-2) - Replacing inductors and capacitors with their dc equivalents

It is easy to find that $I_{1\Omega}=\frac{9V}{2\Omega +1 \Omega}=3A$ and $I_{3mH}=1A$ . Therefore, $I_{2mH}=I_{3mH}+I_{1\Omega}=4A$ , $V_{20\mu F}=-1\Omega \times I_{1\Omega}=-3V$ and $V_{10\mu F}=-2\Omega \times I_{1\Omega}=-6V$ . The total energy stored in the circuit is the sum of the energy stored in elements capable of storing energy, i.e. two capacitors and two inductors. Recall that the energy stored in an inductor is $w_{L}=\frac{1}{2}Li^2_L(t)$ and is equal to $w_{C}=\frac{1}{2}CV^2_C(t)$ for a capacitor. Thus,
$w_{3mH}=\frac{1}{2}(3mH)(1A)^2=1.5\,mJ$
$w_{2mH}=\frac{1}{2}(2mH)(4A)^2=16\,mJ$

$w_{20 \mu F}=\frac{1}{2}(20 \mu F)(-3V)^2=90\,\mu J$
$w_{10 \mu F}=\frac{1}{2}(10 \mu F)(-6V)^2=180\,\mu J$
The total stored energy is $17.77 mJ$ .

Now, switch the sources as shown in Fig. (1-28-3) and calculate the total stored energy. The answer is $60.655 mJ$ . Some of the key quantities are $I_{3mH}=4.5A$ , $I_{2mH}=5.5A$ , $V_{10\mu F}=2V$ and $V_{20\mu F}=1V$ .

Fig. (1-28-3) - Homework

## 12 thoughts on “Total Energy Stored - Circuit with Capacitors and Inductors”

1. thirulok

Now, switch the sources as shown in Fig. (1-28-3) and calculate the total stored energy. The answer is 60.655 mJ . Some of the key quantities are $latex I_{3mH}=4.5A$ , $latex I_{2mH}=5.5A$ , $latex V_{10\mu F}=2V$ and $latex V_{20\mu F}=1V$.

sir, how could we get these answers??

1. Dr. Yaz Z. Li

It is up to you to get those
Please follow the same procedure used to solve the main problem. You should be able to solve this one easily if you understand the main solution. Let me know if you need any help in understanding the main solution.

In the Ohm's law $latex V=RI$, the convention is that the positive terminal of the voltage is the one where the current entering to the element. So, you can assume that $latex V_{1\Omega}=3V$ but $latex V_{20 \mu F}=-V_{1\Omega}=-3V$, because the positive terminal of $latex V_{20 \mu F}$ is the negative terminal of $latex V_{1\Omega}$ and vice versa.