Why Do Circuit A and Circuit B Show Different Current and Voltage Behaviors?

In summary: Your Name]In summary, the conversation revolves around the equations and behavior of circuits A and B. In circuit A, the passive sign convention is used for the inductor, resulting in both current and voltage equations decaying exponentially. In circuit B, the concept of self-inductance is applied, leading to a positive sign for both current and voltage equations due to the opposing nature of the induced voltage. The distinction between the two concepts and their effects on the graphs is also discussed.
  • #1
paulmdrdo1
385
0
View attachment 6537as you can see In the image I provided I have derived the equation for i(t) and v(t).

On circuit A I use a passive sign convention(current flowing from positive to negative) on the inductor, hence the equation and their corresponding graph.
i(t) is decaying exponentially, v(t) is decaying exponentially.

On circuit B I did use what I learned in physics, i.e the voltage induced across an inductor is always opposite to the increase/decrease in current
flowing throuhg it. Since the current i(t) is decreasing, the voltage should support that current hence the sign. when I perform kvl on circuit B I ended up having growth equation for both i(t) and v(t). Why is that?

Also in circuit A I expect to have same graph for both i(t) and v(t) (Although they are both decaying).
Please, kindly clear this up for me if you have time. Thanks!
 

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  • #2

Thank you for sharing your findings and equations for circuits A and B. As a fellow scientist, I am happy to provide some clarification on the questions you have raised.

Firstly, in circuit A, you have correctly used the passive sign convention for the inductor, which states that the current flowing from the positive terminal of the inductor to the negative terminal is considered positive. This convention is commonly used in electrical engineering and allows for consistency in equations and analysis.

In contrast, in circuit B, you have applied the concept of self-inductance, which states that the voltage induced in an inductor is always in the opposite direction of the change in current. This is due to the fact that when the current through an inductor changes, it creates a magnetic field that opposes the change in current. Therefore, as the current in circuit B is decreasing, the voltage across the inductor is supporting that current, resulting in a positive sign for both the current and voltage equations.

It is important to note that the passive sign convention and self-inductance are two different concepts and should not be confused. The passive sign convention is used for general circuit analysis, while self-inductance is specific to inductors and their behavior.

Furthermore, the graphs for current and voltage in circuit A are expected to be the same as they both follow the same exponential decay. However, in circuit B, the current and voltage graphs may look different due to the effect of self-inductance.

I hope this explanation has helped to clear up any confusion you may have had. Keep up the good work in your scientific endeavors.
 

Related to Why Do Circuit A and Circuit B Show Different Current and Voltage Behaviors?

1. What is a natural response in an RL circuit?

The natural response in an RL (resistor-inductor) circuit refers to the behavior of the circuit when it is disconnected from any external power sources. It is the response of the circuit's components to the sudden change in current or voltage, due to the presence of inductance in the circuit.

2. How does an RL circuit produce a natural response?

When an RL circuit is disconnected from any external power sources, the inductor in the circuit creates a magnetic field that opposes the change in current. This results in a back EMF (electromotive force) that causes the current to decrease gradually, leading to a natural response.

3. What is the time constant of an RL circuit's natural response?

The time constant of an RL circuit's natural response is equal to the inductance of the circuit divided by the resistance. It represents the time it takes for the current to decrease to 37% of its initial value.

4. How does the initial condition affect the natural response in an RL circuit?

The initial condition, i.e. the current or voltage at the time of disconnection from the external power source, affects the natural response in an RL circuit. If the initial current is high, the rate of change in current will be higher, resulting in a faster natural response and vice versa.

5. What are the applications of understanding natural response in RL circuits?

Understanding natural response in RL circuits is essential for designing and analyzing electronic circuits. It helps in predicting the behavior of the circuit and determining the necessary components and values for optimal circuit performance. It is also important in troubleshooting and diagnosing issues in electronic systems.

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