Total Energy Stored – Circuit with Capacitors and Inductors

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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

capacitor dc analysis Energy inductor passive elements stored energy storing energy terminal characteristics

Comments

13 responses to “Total Energy Stored – Circuit with Capacitors and Inductors”

December 30, 2010

Kalialberts Appah

I am working on a project ‘using teaching and learning
materials to teach inductors and capacitors in electric
circuits

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December 30, 2010

Kalialberts Appah

please i need your help

Reply
1. December 30, 2010
  
  Dr. Yaz Z. Li
  
  How can I help you?
  
  Reply
  1. November 24, 2013
    
    can
    
    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
January 10, 2011

myun

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March 20, 2011

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 $I_{3mH}=4.5A$ , $I_{2mH}=5.5A$ , $V_{10\mu F}=2V$ and $V_{20\mu F}=1V$ .

sir, how could we get these answers??

Reply
1. March 22, 2011
  
  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.
  
  Reply
October 23, 2011

Qasim Muhammad

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

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December 12, 2011

janebi5

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

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January 19, 2012

Sardeep

Very nicely explained. plain ans simple

Reply
February 17, 2012

Dixon

Why is the negative sign used while computing V20microF?

Reply
1. February 18, 2012
  
  Dr. Yaz Z. Li
  
  In the Ohm’s law $V=RI$ , 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 $V_{1\Omega}=3V$ but $V_{20 \mu F}=-V_{1\Omega}=-3V$ , because the positive terminal of $V_{20 \mu F}$ is the negative terminal of $V_{1\Omega}$ and vice versa.
  
  Reply
May 29, 2012

Ema watson

i need more solved problms how could i get those????

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