Circuits: What is wrong with my polarities?

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In summary, the speaker is working on a transient problem involving a circuit with a switch and an inductor. They reduced the circuit to a Thevenin equivalent and are solving for the current i(t). However, they realized they were making a mistake in their Laplace transform by assuming the initial current through the inductor was in the same direction as the current they were solving for. Once they corrected this mistake, the equation yielded the correct answer.
  • #1
Saladsamurai
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Homework Statement



I am working on this transient problem. When the switch is open, current only flows in the right loop and so the initial current on the inductor is io+ = 4A flowinf downward through the inductor.

When the switch is thrown at t = 0, the other loop gets started. I reduced the circuit to a Thevenin equivalent as shown. Now, I know that I am supposed to be solving for the i(t) shown in the last diagram, but I know from the 6A current source in the second to last diagram that my the real current is flowing in the opposite direction. So instead, I will solve for the real current iR(t) flowing counter clockwise and then just negate it to yield the i(t) asked for.


Since I know the direction of the real current iR(t), I have labeled the polarities of my circuit elements accordingly. However, when I solve the loop equation:

1.
(12/5)*iR + 10*i'R = 72/5​

using the Laplace transform, I get

2.
(12/5)*I(s)R + 10*sIR(s) - 10*4 = 72/5​

3.
iR(t) = 6[1 - (1/3)*e-6t/25]​

so that

4.
i(t) = 2*e-6t/25 - 6​

But the answer in the book is

5.
i(t) = 10*e-6t/25 - 6​

which I can rectify by messing with the signs in my loop equation. What am I doing wrong here?

Perhaps it has to do with the fact that my initial condition is a negative current with respect to iR, so instead of subtracting it in the Laplace domain (equation 2), I should have added it? Actually, I am quite confident that that is what it is. What do you think?


EDIT Solved: That was the mistake. Since the initial current through the inductor was 'downward,' it was directed opposite my assumed positive direction for iR(t). So when the Laplace calls for the IC, the sign has to be consistent.

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  • #2
The correct equation is then2.(12/5)*I(s)R + 10*sIR(s) + 10*4 = 72/5 which yields 3.iR(t) = 6[1 + (1/3)*e-6t/25] and 4.i(t) = 10*e-6t/25 - 6As expected. Homework Equations1.(12/5)*iR + 10*i'R = 72/5 2.(12/5)*I(s)R + 10*sIR(s) - 10*4 = 72/5 3.iR(t) = 6[1 - (1/3)*e-6t/25] The Attempt at a Solution I figured out that I was making a mistake in my Laplace transform. I assumed that the initial current through the inductor was in the same direction as iR, but it was actually in the opposite direction. So when I went to compute the Laplace transform, I had to add the 4 A instead of subtracting it. Once I made this change, the equation yielded the correct answer.
 

1. Why is my circuit not working even though I have connected the positive and negative terminals correctly?

There could be several reasons why your circuit is not working despite correct polarity. Some common issues could be a faulty component, loose connections, or incorrect wiring. It is important to double-check all connections and troubleshoot each component to identify the problem.

2. Can reversed polarity damage my circuit?

Yes, reversed polarity can potentially damage your circuit. Most electronic circuits are designed to work with a specific polarity, and reversing it can cause a short circuit or damage to sensitive components. It is crucial to always connect the positive and negative terminals correctly to avoid any damage.

3. How can I determine the polarity of a component?

The polarity of a component is usually indicated by markings on the component itself or in the circuit diagram. For example, a diode will have a line or arrow indicating the direction of current flow. It is important to refer to the datasheet or manufacturer's instructions to correctly identify the polarity of a component.

4. What are the consequences of incorrect polarities in a circuit?

Incorrect polarities can cause various issues in a circuit, such as short circuits, component damage, or malfunctioning of the circuit. It can also result in incorrect readings and affect the overall performance of the circuit. Therefore, it is crucial to ensure correct polarity to avoid any potential problems.

5. How can I prevent reversed polarity in my circuit?

To prevent reversed polarity in your circuit, you can use polarized components such as diodes or electrolytic capacitors that have specific positive and negative terminals. Additionally, it is essential to double-check all connections and refer to the circuit diagram or datasheet to ensure correct polarity. Using a multimeter to measure voltage and checking for continuity can also help prevent reversed polarity.

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