Solving a Node Analysis Problem with 0.632V Voltage Source

In summary, the conversation discusses a circuit problem and the use of node analysis to solve for the voltage across an inductor. The initial current in the inductor is determined to be -0.126 A and the voltage across it is found to be 0 V at t = 0+. There is some confusion about why the voltage is zero, but it is concluded that the current will decay over time due to the resistors in the circuit. The solution manual may have made an error in using a constant current instead of a function to find the voltage at t = 0+.
  • #1
dwn
165
2

Homework Statement



Attached image


Using node analysis:
ix(0-) = (v-4)/10 + v/3 + v/5 = 0
v = 0.632

ix(0-) = -v/3 = -0.211 A
I don't understand why they didn't use a current divider since there is a 5 ohm resistor in parallel with the 3 ohm. Current changes for resistors that are in parallel...

il(0-) = v/5 = -0.63/5 = 0.126A

Then when it comes to solve for v(0+) = di/dt = d(0.126)/dt = 0
How can the voltage across an inductor be zero when it's charged by the 4V source for t < 0??
 

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  • #2
dwn said:

Homework Statement



Attached image


Using node analysis:
ix(0-) = (v-4)/10 + v/3 + v/5 = 0
v = 0.632

ix(0-) = -v/3 = -0.211 A
I don't understand why they didn't use a current divider since there is a 5 ohm resistor in parallel with the 3 ohm. Current changes for resistors that are in parallel...
They could have, but it would be as much or more work since you need the total current to begin with (or make a Norton equivalent out of the 4V and 10 Ω resistor and have a three leg current divider). Nodal analysis is pretty straightforward and doesn't require any electronic gymnastics to apply to the circuit as given.

il(0-) = v/5 = -0.63/5 = 0.126A

Then when it comes to solve for v(0+) = di/dt = d(0.126)/dt = 0
How can the voltage across an inductor be zero when it's charged by the 4V source for t < 0??
Why do you say that di/dt = 0?

When the switch opens the inductor starts with some initial current. The inductor will produce whatever EMF is required to try to maintain that current in the first instant. But the current will decay from there (and so will the EMF) as the resistors bleed away energy (heat).
 
  • #3
gneill said:
Why do you say that di/dt = 0?

When the switch opens the inductor starts with some initial current. The inductor will produce whatever EMF is required to try to maintain that current in the first instant. But the current will decay from there (and so will the EMF) as the resistors bleed away energy (heat).

Exactly, that's what I was thinking, but the solution manual says otherwise. I think they made the error and used the constant iL(0) = 0.126A instead of the function 0.126e^(-t/tau) when trying to find vL(0+). Safe assessment?
 
  • #4
dwn said:
Exactly, that's what I was thinking, but the solution manual says otherwise. I think they made the error and used the constant iL(0) = 0.126A instead of the function 0.126e^(-t/tau) when trying to find vL(0+). Safe assessment?

Dunno. But it is certainly true that for the instant t = 0+ the current will be the same as the current at t = 0-. That is, it will be 0.126 A for that instant and can be treated as a constant value for the voltage calculation at t = 0+.
 
  • #5


I would like to clarify a few things in this problem. First, it is important to note that the given circuit is not complete and lacks some essential components such as a ground connection and a load. These missing elements can significantly affect the solution and must be considered when solving the problem.

Regarding the use of node analysis, it is a commonly used method in circuit analysis to determine the voltage and current at different nodes in a circuit. In this case, the node analysis equation is correctly set up, and the solution for v is also correct. However, I agree that using a current divider could also be a valid approach in this problem, but it may not provide the same level of accuracy as node analysis.

Moving on to the second part of the problem, it is important to understand that the voltage across an inductor can indeed be zero, even when it is charged by a voltage source. This is because an inductor resists changes in current, and when the current through it is constant, the voltage across it becomes zero. In this problem, the inductor is initially charged by the 4V source, but as time passes, the current through it becomes constant, and the voltage across it drops to zero.

In conclusion, it is essential to have a complete understanding of the given circuit and its components before attempting to solve a problem. Additionally, different methods can be used to solve a circuit, but the accuracy and appropriateness of each method must be carefully evaluated.
 

1. What is node analysis and why is it useful in solving circuit problems?

Node analysis is a method used to analyze electrical circuits by dividing them into nodes, or specific points where the circuit branches meet. It is useful because it allows us to determine the voltage and current at each node, which is crucial in understanding how a circuit functions and how to troubleshoot any issues that may arise.

2. How does a 0.632V voltage source affect the node analysis process?

A 0.632V voltage source, also known as a "Thevenin equivalent", can simplify the node analysis process by replacing the original circuit with a single voltage source and a resistor. This makes it easier to calculate the voltage and current at each node.

3. Can a node analysis problem with a 0.632V voltage source have multiple solutions?

Yes, it is possible for a node analysis problem to have multiple solutions. This can occur when the circuit contains dependent sources, such as voltage-dependent or current-dependent sources. In these cases, the solution will depend on the values of the dependent sources.

4. What are some common mistakes to avoid when solving a node analysis problem with a 0.632V voltage source?

Some common mistakes to avoid include forgetting to include all the necessary components in the node equations, not properly labeling the nodes, and making calculation errors. It is also important to double-check the final solution to ensure it is physically and logically possible.

5. How can I check if my solution to a node analysis problem with a 0.632V voltage source is correct?

The best way to check your solution is to simulate the circuit using a software or online simulator, such as LTSpice or CircuitLab. This will allow you to compare the simulated results to your calculated values and identify any discrepancies. Additionally, you can also verify your solution by using KCL and KVL equations to ensure they are satisfied at each node in the circuit.

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