Find the total energy stored in the circuit.

find the 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).
Energy stored in circuit - dc equivalent
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 .

Find the energy stored in circuit - Homework
Fig. (1-28-3) - Homework

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13 Comments

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      1. hi dr. yaz z.li
        I have an UPS 40kVA power of two connected in series with the inductor burned. I was wrapped up, but warming up again. may the toroid core be defective thank you so much in advance

        Reply
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  4. 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??

    Reply

    1. 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.

      Reply

  5. Respected Professor,
    I am thankful to you. its really helpful to me.may Allah (God) bless you and give you more strength to help and guide the students,My wishes to you.Thanks

    Reply

  6. Hello!
    How can I upload image with my problem?Actually this problem is connected with this theme.

    Reply
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  8. Reply

    1. 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.

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