Calculating Instantaneous Current in an Inductive Circuit with Delayed Switches

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Homework Help Overview

The discussion revolves around calculating the instantaneous current in an inductive circuit where two switches are involved, with one switch closing after a delay. The subject area includes concepts from electrical circuits, specifically dealing with inductance, resistance, and current flow over time.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of time constants and maximum current in the circuit when different switches are closed. There are attempts to derive expressions for current at various times and to understand the behavior of the circuit before and after the switches are closed. Questions arise regarding the initial conditions and how they affect subsequent calculations.

Discussion Status

The discussion is active, with participants providing various calculations and questioning each other's approaches. Some participants suggest re-evaluating the initial current before switch S1 closes and how it influences the current after S1 is closed. There is an ongoing exploration of the exponential growth and decay of current in the circuit.

Contextual Notes

Participants note the importance of considering the initial current from the circuit configuration before S1 closes, as well as the final steady-state current when S1 is closed. There are references to specific resistance values and the impact of closing switches on the circuit's behavior.

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Homework Statement


In the inductive circuit shown in Figure 2, the switch S1 is closed 0.75s after switch S2 is closed. Calculate the instantaneous current in the coil 1.5s after switch S1 is closed.
upload_2016-3-27_13-10-3.png


Homework Equations


T=L/R
I=V/R

The Attempt at a Solution


When S2 is closed:
T=L/R
=2.4/7.2
=0.333
=333ms

IMAX=V/R
=24/7.2
=3.33A

At time 0.75s S1 is closed, the current is:
i1=IMAX(1-e^-t/T)
=24/3.6(1-e^-750/333)
=5.97A

When S1 is closed current flows through the 3.6ohm resistor and decays:
Time constant is:
T=L/R
=2.4/3.6
=0.666
=666ms

Current after 1.5s after S1 is closed:
i2=i1e^-t/T
=5.97e^-(1.5x10^-3/666)
=5.97A

This cannot be correct?
 
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DevonZA said:
When S2 is closed:
T=L/R
=2.4/7.2
=0.333
=333ms
What's the total resistance in the series circuit when only S2 is closed?
 
gneill said:
What's the total resistance in the series circuit when only S2 is closed?

7.2ohms+3.6ohms = 10.8ohms

Paris_Tuileries_Garden_Facepalm_statue.jpg
 
I get an answer of 6.44A for i1 and i2
 
DevonZA said:
I get an answer of 6.44A for i1 and i2
That doesn't look right to me. Can you show some details?

What was the current just before S1 closed?
After S1 closes, what final current is the circuit heading for?
 
When S2 is closed:
T=L/R
=2.4/7.2+3.6
=0.222
=222ms

IMAX=V/R
=24/7.2+3.6
=2.22A

At time 0.75s S1 is closed, the current is:
i1=IMAX(1-e^-t/T)
=24/3.6(1-e^-750/222)
=6.44A

When S1 is closed current flows through the 3.6ohm resistor and decays:
Time constant is:
T=L/R
=2.4/3.6
=0.666
=666ms

Current after 1.5s after S1 is closed:
i2=i1e^-t/T
=6.44e^-(1.5x10^-3/666)
=6.44A
 
DevonZA said:
When S2 is closed:
T=L/R
=2.4/7.2+3.6
=0.222
=222ms

IMAX=V/R
=24/7.2+3.6
=2.22A

At time 0.75s S1 is closed, the current is:
i1=IMAX(1-e^-t/T)
=24/3.6(1-e^-750/222)
=6.44A
Why didn't you use the IMAX value that you found? You've dropped the 7.2 Ω resistor again.

Edit: Sorry, I just realized that you were writing an expression for I(t) for the second circuit and that you're implying a new IMAX. Two problems though, you used the same time constant as for the previous circuit configuration (it should change when the total resistance changes), and you didn't take into account the initial current that the circuit starts with at the instant S1 closed.
When S1 is closed current flows through the 3.6ohm resistor and decays:
Why does it decay? What is the final current that the circuit is heading for with a 24 V supply and 3.6 Ω? Is this current smaller or greater than the initial current (at the instant S1 closes)?
 
Last edited:
gneill said:
Why didn't you use the IMAX value that you found? You've dropped the 7.2 Ω resistor again.

Edit: Sorry, I just realized that you were writing an expression for I(t) for the second circuit and that you're implying a new IMAX. Two problems though, you used the same time constant as for the previous circuit configuration (it should change when the total resistance changes), and you didn't take into account the initial current that the circuit starts with at the instant S1 closed.

Why does it decay? What is the final current that the circuit is heading for with a 24 V supply and 3.6 Ω? Is this current smaller or greater than the initial current (at the instant S1 closes)?

Final current is I=V/R 24/3.6=6.66A the 7.2ohm resistor is bypassed when S1 is closed? The current is greater because when S1 is open we pass through both the 7..2ohm and 3.6ohm resistors.

Which step looks wrong to you? Am I even headed in the right direction with my attempt?
 
You haven't determined the current in the circuit for the instant *before* S1 closes. That current is the result of the the initial configuration (with both resistors) running for 0.75 seconds. That current is also the initial current that the new configuration starts with when S1 closes. You need to take that initial current into account.
 
  • #10
gneill said:
You haven't determined the current in the circuit for the instant *before* S1 closes. That current is the result of the the initial configuration (with both resistors) running for 0.75 seconds. That current is also the initial current that the new configuration starts with when S1 closes. You need to take that initial current into account.
I=V/R
I=24/10.8
=2.22A

AT 0.75S:
i1=24/10.8(1-e^-(750/222)
= 2.15A
 
  • #11
DevonZA said:
I=V/R
I=24/10.8
=2.22A

AT 0.75S:
i1=24/10.8(1-e^-(750/222)
= 2.15A
Yup. You might want to keep a couple more digits as this is an intermediate step value that will be used for further calculations. Only round final results for presentation.

Now, how are you going to use this as the initial current for the next step (S1 closes)?
 
  • #12
gneill said:
Yup. You might want to keep a couple more digits as this is an intermediate step value that will be used for further calculations. Only round final results for presentation.

Now, how are you going to use this as the initial current for the next step (S1 closes)?

Thank you for your help thus far :)

Okay how does this look:
When S2 is closed:
T=L/R
=2.4/10.8
=0.222
=222ms

IMAX=V/R
=24/10.8
=2.22A

At time 0.75s S1 is closed, the current is:
i1=IMAX(1-e^-t/T)
=24/10.8(1-e^-750/222)
=2.15A

When S1 is closed current flows through the 3.6ohm resistor:
Time constant is:
T=L/R
=2.4/3.6
=0.666
=666ms

Current after 1.5s after S1 is closed:
i2=i1e^-t/T
=2.15e^-(1500/666) Note: 1.5 seconds = 1500ms right?
=0.23A
 
  • #13
DevonZA said:
AT 0.75S:
i1=24/10.8(1-e^-(750/222)
= 2.15A
Right. Now from this value, the current increases exponentially to 6.66A when S1 is closed. Write an equation which gives you i=2.15A for t=0 and i=6.66A for t=∞.(assuming S1 is closed at t=0)
 
Last edited:
  • #14
DevonZA said:
Current after 1.5s after S1 is closed:
i2=i1e^-t/T
=2.15e^-(1500/666) Note: 1.5 seconds = 1500ms right?
=0.23A
The current will start at ~ 2.15A, but where will it be headed? What would be the final current value after a long time has passed?
 
  • #15
DevonZA said:
Current after 1.5s after S1 is closed:
i2=i1e^-t/T
=2.15e^-(1500/666) Note: 1.5 seconds = 1500ms right?
=0.23A
How will the current decay? Check your equation for i2.
Assume switch S1 is closed at t=0 and initial current as 2.15A.
 
  • #16
Final current is I=V/R 24/10.8=2.22A?
So current increases..
I'm confused?
 
  • #17
DevonZA said:
Final current is I=V/R 24/10.8=2.22A?
So current increases..
I'm confused?
No, for this part switch S1 is closed. What is the steady state current for this configuration?
 
  • #18
DevonZA said:
Final current is I=V/R 24/10.8=2.22A?
So current increases..
I'm confused?
DevonZA said:
Final current is I=V/R 24/3.6=6.66A the 7.2ohm resistor is bypassed when S1 is closed? The current is greater because when S1 is open we pass through both the 7..2ohm and 3.6ohm resistors.
You've got it right already..
 
  • #19
gneill said:
No, for this part switch S1 is closed. What is the steady state current for this configuration?
Sorry that is my mistake I should have written I=V/R 24/3.6=6.66A

Okay so initial current is 2.15A and final current is 6.66A

How do I find instantaneous current in the coil 1.5s after S1 closes?
 
Last edited:
  • #20
So you know that the current will start at 2.15 A and head towards 6.67 A, following an exponential curve. How might you write an expression for this?
 
  • #21
gneill said:
So you know that the current will start at 2.15 A and head towards 6.67 A, following an exponential curve. How might you write an expression for this?
http://www.regentsprep.org/regents/math/algebra/ae7/ExpDec13.gifa = initial amount before measuring growth/decay
r = growth/decay rate (often a percent)
x = number of time intervals that have passed

y=2.15(1+2.15/6.67)^1.5
=3.67A
 
  • #22
No, that's not the right approach. Stick to the curves for the exponential function associated with the circuit.

What is the total amount that the current can grow from initial to steady state? (i.e. what's the ##ΔI##?)
 
  • #23
ΔI=6.67-2.15=4.52A
 
  • #24
Right. So the current will be tending to grow by 4.52 A over time. It will follow an exponential curve as you've used in previous parts. 4.52 A will be the magnitude (your "IMAX") for that curve. This growth sits on top of the initial current. So:
$$I(t) = I_o + ΔI(1 - e^{-t/\tau})$$
 
  • #25
gneill said:
Right. So the current will be tending to grow by 4.52 A over time. It will follow an exponential curve as you've used in previous parts. 4.52 A will be the magnitude (your "IMAX") for that curve. This growth sits on top of the initial current. So:
$$I(t) = I_o + ΔI(1 - e^{-t/\tau})$$

It=2.15+4.52(1-e^-(1500/666)
=6.19A
 
  • #26
That looks good.
 
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  • #27
Thank you gneill and cnh1995 truly appreciate your help :)
If you are still in the mood for assisting I could use some help with my reluctance question.
 
  • #28
Hi Devon

Can you please show your end result , I am also struggling with this.
 

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