Calculating Instantaneous Current in an Inductive Circuit with Delayed Switches

[DevonZA]

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.

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?

 

[gneill]

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?

 

[DevonZA]

What's the total resistance in the series circuit when only S2 is closed?

7.2ohms+3.6ohms = 10.8ohms

 

[DevonZA]

I get an answer of 6.44A for i1 and i2

 

[gneill]

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?

 

[DevonZA]

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

 

[gneill]

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)?

 

[DevonZA]

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?

 

[gneill]

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.

 

[DevonZA]

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

 

[gneill]

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)?

 

[DevonZA]

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

 

[cnh1995]

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)

 

[gneill]

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?

 

[cnh1995]

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.

 

[DevonZA]

Final current is I=V/R 24/10.8=2.22A?
So current increases..
I'm confused?

 

[gneill]

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?

 

[cnh1995]

Final current is I=V/R 24/10.8=2.22A?
So current increases..
I'm confused?

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..

 

[DevonZA]

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?

 

[gneill]

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?

 

[DevonZA]

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

 

[gneill]

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 $ΔI$?)

 

[DevonZA]

ΔI=6.67-2.15=4.52A

 

[gneill]

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:
$$I(t) = I_o + ΔI(1 - e^{-t/\tau})$$

 

[DevonZA]

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

 

[gneill]

That looks good.

 

[DevonZA]

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.

 

[Brandon de Jager]

Hi Devon

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