Calculating Instantaneous Current in an Inductive Circuit with Delayed Switches

In summary, in the given inductive circuit, switch S1 is closed 0.75s after switch S2 is closed. The instantaneous current in the coil can be calculated by using the equations T=L/R and I=V/R. When S2 is closed, the time constant is 333ms and the maximum current is 3.33A. When S1 is closed, the current initially decays with a time constant of 666ms and then increases exponentially to a final current of 6.66A. The initial current is 2.15A and the final current after a long time has passed is 6.66A. The current decays according to the equation i2=i1e^-t
  • #1
DevonZA
181
6

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?
 
Physics news on Phys.org
  • #2
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?
 
  • #3
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
 
  • #4
I get an answer of 6.44A for i1 and i2
 
  • #5
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?
 
  • #6
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
 
  • #7
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:
  • #8
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?
 
  • #9
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.
 
  • Like
Likes DevonZA
  • #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.
 

Related to Calculating Instantaneous Current in an Inductive Circuit with Delayed Switches

What is instantaneous current in a coil?

Instantaneous current in a coil refers to the flow of electric charge through a coil at a given moment in time. It is measured in amperes (A) and represents the rate at which electric charge is moving through the coil.

How is instantaneous current in a coil different from average current?

Instantaneous current refers to the current at a specific point in time, whereas average current is the overall flow of electric charge over a period of time. Instantaneous current can vary greatly, while average current represents a more stable measure of the flow of electric charge.

What factors affect the magnitude of instantaneous current in a coil?

The magnitude of instantaneous current in a coil can be affected by factors such as the voltage applied to the coil, the resistance of the coil, and the inductance of the coil. Changes in these factors can result in fluctuations in the instantaneous current.

How is instantaneous current measured in a coil?

Instantaneous current in a coil can be measured using an ammeter, which is a device designed specifically to measure electric current. The ammeter is connected in series with the coil, and the current passing through the coil is then displayed on the device.

Why is it important to understand instantaneous current in a coil?

Understanding instantaneous current in a coil is important for various reasons. It allows us to analyze the behavior of electrical circuits and design more efficient systems. It is also crucial for safety purposes, as high levels of instantaneous current can cause damage to electrical components and pose a risk to individuals working with them.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
307
  • Introductory Physics Homework Help
Replies
8
Views
379
  • Introductory Physics Homework Help
Replies
34
Views
6K
Replies
3
Views
8K
  • Introductory Physics Homework Help
Replies
16
Views
2K
Replies
6
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
6K
  • Introductory Physics Homework Help
Replies
5
Views
2K
Back
Top