What is the maximum current in a RL circuit using V=IR?

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Homework Help Overview

The discussion revolves around determining the maximum current in a series RL circuit with given resistance, inductance, and voltage values. Participants are exploring the implications of the time constant and the behavior of current over time after the switch is closed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the time constant and initial current, questioning the accuracy of their values and the assumptions made about the circuit's behavior. There are inquiries about the correct interpretation of the equations used and the significance of the switch closing time.

Discussion Status

The conversation is ongoing, with participants providing hints and guidance to each other. Some express frustration, while others encourage a focus on understanding the underlying concepts rather than just applying formulas. There is a recognition that the maximum current can be derived from basic principles without complex calculations.

Contextual Notes

Participants are grappling with the definitions of initial and maximum current in the context of RL circuits, as well as the relevance of the time constant in their calculations. There is a suggestion that some values and concepts may be distractions from the core question.

skibidi
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Homework Statement
The switch in a series RL circuit with a resistance of 5.8 Ω , inductance of 3.6 H, and
voltage of 23.7 V is closed at t = 0.5 s.
What is the maximum current in the circuit?
Answer in units of A.
Relevant Equations
I(f) = I(1-e^(-t/T))
I found the time constant using L/R and got 0.62s (3.6H/5.8ohms)
I found the initial current V/R and got 4.086A (23.7V/5.8ohms)

I(f) = 4.086A(1-e^(-0.5s/0.62s))
Then I plugged it into the equation and got 2.26A and it was wrong
 
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The exponential current response starts when the switch is closed at t=0.5 sec. That piece of information doesn't appear in your equation. Would you expect the same equation if it closed at t=1 sec, t= 100sec?

I suggest you draw a graph of that equation and compare it to your intuition about how the circuit will work.
 
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skibidi said:
Homework Statement: The switch in a series RL circuit with a resistance of 5.8 Ω , inductance of 3.6 H, and
voltage of 23.7 V is closed at t = 0.5 s.
What is the maximum current in the circuit?
Answer in units of A.
Relevant Equations: I(f) = I(1-e^(-t/T))

I found the initial current V/R and got 4.086A (23.7V/5.8ohms)
4.086 A isn't the initial current
 
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I found that to be the maximum current
 
Would I use the equation I used to solve for the current at a given time such as 1s.
 
The equation you wrote assumes the switch closes at t=?
 
0.5
 
So would t be the duration since 0.5s, Some t value - 0.5s

Let's say that I had to find the current at 1.1s, so would use the equation and substitute t for 0.6s.

Would it be either
I = 4.09A(1-e^(-1.1s/(L/R))
or I = 4.09A(1-e^(-0.6s/(L/R))

The second one was correct
 
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I hate this so much...
 
  • #10
Sketch a graph of the waveform ##v(t)=1-e^\frac{-t}{T}## versus ##t## where ##T## is a constant, like 0.6 sec. What (where) is it's maximum value?
 
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  • #12
skibidi said:
I hate this so much...
1. Don't give up! You are making a very simple mistake.

skibidi said:
Relevant Equations: I(f) = I(1-e^(-t/T))
2. How's your maths? Suppose the switch is closed at t=0. Can you look at this equation and say how big the current is at t=0? And can you say how big the current is a long time later (when t is very large)?

skibidi said:
I found the time constant using L/R and got 0.62s (3.6H/5.8ohms)
Ok. But the irony is that you don't need the time constant to answer the question! (You don't even need the value of L or the time at the switch is closed! They are 'red herrings'.)

skibidi said:
I found the initial current V/R and got 4.086A (23.7V/5.8ohms)
3. The initial current is not 4.086A. (If you were using a capacitor rather than an inductor, you would be correct. Maybe that was in the back of your mind.) With the inductor, what is the current the moment the switch is closed? (Hint - see item 2 above).

skibidi said:
I(f) = 4.086A(1-e^(-0.5s/0.62s))
Then I plugged it into the equation and got 2.26A and it was wrong
4. It should have been obvious that something was wrong!. 2.26A is smaller tha 4.086A. So 2.26A could not be the maximum (biggest) current!

5. Go back and read through the replies. @DaveE and @Gordianus were giving big hints. The question is really easy (no complicated maths) once you understand what's going on. Let us know if you're still stuck.
 
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  • #13
skibidi said:
I hate this so much...
Why are you focused on the time constant? That's just a value of ##t## for a particular value of ##I##.

I suspect you are attempting success in the course using an answer-making strategy. But the entire purpose of the course is to get you to develop sense-making strategies.

This is the disconnect that's the cause of your frustration. Try to make sense of what's happening in that circuit, then the answer will come to you easily.

Develop that skill and you'll be successful in the course without those negative emotions like frustration and hate.
 
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  • #14
For max current you literally only have to use the V=IR formula.
 

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