Total Energy Stored – Circuit with Capacitors and Inductors

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.
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Thévenin’s Theorem – Circuit with Two Independent Sources

Use Thévenin’s theorem to determine I_O.

Thevenin's Theorem - Circuit containing two independent sources

Fig. (1-27-1) - Circuit with two independent sources


Solution
Lets break the circuit at the 3\Omega load as shown in Fig. (1-27-2).
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Thévenin’s Theorem – Circuit with An Independent Source

Use Thévenin’s theorem to determine V_O .

A circuit with a voltage source

Fig. (1-26-1) - The Circuit

Solution
To find the Thévenin equivalent, we break the circuit at the 4\Omega load as shown below.
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Superposition Method – Circuit With Dependent Sources

Determine I_x , I_y and V_z using the superposition method.

Superposition - Circuit with dependent sources
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Superposition Problem with Four Voltage and Current Sources

Determine V_x and I_x using the superposition method.
A circuit with four voltage and current sources to be solved by the superposition method
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Turning Sources Off

Turning off a source, which is usually used in solving circuits with superposition method, means setting its value equal to zero. For a voltage source, setting the voltage equal to zero means that it produces zero voltage between its terminals. Therefore, the voltage source must insure that the voltage across two terminals is zero. Replacing the source with a short circuit can do that. Thus, voltage sources become a short circuit when turned off.

For a current source, setting the current equal to zero means that it produces zero current. Therefore, the current source must insure that no current flows through its branch. An open circuit can do that. Hence, to turn off a current source it should be replaced by an open circuit.

How about dependent sources? The voltage/current of a dependent source is dependent on other variables of the circuit. Therefore, dependent sources cannot be turned off.

Example I: Turn off sources one by one.

turning sources off example 1-1

Example 1


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Electrical Circuits eBooks

Nodal Analysis eBook

Nodal Analysis eBook
You are welcome to download our free ebook about Nodal Analysis from here. You need to sign up for our site before downloading. Please note that we will never sell, trade or share your email with a third-party. Occasionally, we will send you an email about new posts, ebooks and exciting things happening in solved-problems.com.

content

  • Introduction
  • Reference Node
  • Node Voltages
  • Nodal Analysis Steps
  • Complicated Cases
    • Circuits with Non-grounded Voltage Sources
    • Circuits with Dependent Current Sources
    • Circuits with Dependent Voltage Sources
  • More Problems
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Nodal Analysis

Nodal Analysis eBook

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Nodal Analysis Steps

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Nodal Analysis – Circuit with Dependent Voltage Source

Determine the power of each source after solving the circuit by the nodal analysis.
Nodal Analysis - Supernode - Dependent Voltage Source 1

Answers: P_{I_x}=0.497W, P_{1A}=-1.806W, P_{2A}=4.254W, P_{3V}=-3.87W, and P_{5V}=-3.552W

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Nodal Analysis – 6-Node Circuit

Determine the power of each source after solving the circuit by the nodal analysis.

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