Rate Of Change Of Current In An Inductor

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  • #1

Homework Statement



A coil has self inductance L of 2 Henrys, and a resistance R of 100 Ohms. A DC supply voltage E of 100 volts is applied to the coil.

Show graphically the approximate rate at which the current increases at the moment of switching on (t=0) and after 10ms.

Confirm your results by differentiation.

Homework Equations



Instantaneous current: i=(E/R)(1-e^-Rt/L)
Slope of a tangent: y2-y1/x2-x1
di/dt: i(b)-i(a)/t(b)-t(a)

The Attempt at a Solution



Sorry guys, would not normally ask for help but I am stuck here and we have broken up for Xmas so I cannot ask anyone at the college I go to. Anyway if someone could have a quick look over this and point me in the right direction I would appreciate it.

* I worked out the time constant to be L/R = 20ms

* I worked out the maximum circuit current to be E/R = 1A

* I created an exponential growth curve showing the inductor current as a function of time using the instantaneous current equation above and taking points at every ms to 15ms then every 5 to 45ms (to show at least 2 time constants on the curve)

* Using the curve I plotted a tangent line across the point at 10ms. I took two points and used the slope equation y2-y1/x2-x1 to calculate the approximate slope at 10ms (30.8A/s)

* I took point A at 10ms and point B at 15ms. I took the co-ordinates for A & B and worked out the average rate of current change between the two points as di/dt = 26.8A/s. I then moved B closer to A and re-calculated. I did this for lesser and lesser values of B and created a table. As B approached A I could confidently predict through di/dt that that A would equal 30A/s at the point of 10ms.

My questions....

1) Can I assume that the simple operation of dy/dx above is enough in regards to differentiation to confirm the graphical approximation I made before it? Or am I looking for another method? (Relating to the 'confirm your results by differentiation' comment in the question)

2) Its asks for the rate of change of current at t=0. I am aware that there is no current through the inductor at t=0 due to the inductor opposing the current change (back emf etc) but there IS a current rate of change at t=0 and it can be calculated as: initial di/dt = E/L Amps. If I use this I get 50A/s as an initial current rate of change. How can this be shown graphically on the growth curve and is the above initial di/dt equation enough to satisfy the 'confirm by differentiation' statement in the question?

Thanks in advance.
Simon
 
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Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Homework Statement



A coil has self inductance L of 2 Henrys, and a resistance R of 100 Ohms. A DC supply voltage E of 100 volts is applied to the coil.

Show graphically the approximate rate at which the current increases at the moment of switching on (t=0) and after 10ms.

Confirm your results by differentiation.

Homework Equations



Instantaneous current: i=(E/R)(1-e^-Rt/L)
Slope of a tangent: y2-y1/x2-x1
di/dt: i(b)-i(a)/t(b)-t(a)

The Attempt at a Solution



Sorry guys, would not normally ask for help but I am stuck here and we have broken up for Xmas so I cannot ask anyone at the college I go to. Anyway if someone could have a quick look over this and point me in the right direction I would appreciate it.

* I worked out the time constant to be L/R = 20ms

* I worked out the maximum circuit current to be E/R = 1A

* I created an exponential growth curve showing the inductor current as a function of time using the instantaneous current equation above and taking points at every ms to 15ms then every 5 to 45ms (to show at least 2 time constants on the curve)

* Using the curve I plotted a tangent line across the point at 10ms. I took two points and used the slope equation y2-y1/x2-x1 to calculate the approximate slope at 10ms (30.8mA/s)

* I took point A at 10ms and point B at 15ms. I took the co-ordinates for A & B and worked out the average rate of current change between the two points as di/dt = 26.8mAs. I then moved B closer to A and re-calculated. I did this for lesser and lesser values of B and created a table. As B approached A I could confidently predict through di/dt that that A would equal 30mA/s at the point of 10ms.

My questions....

1) Can I assume that the simple operation of dy/dx above is enough in regards to differentiation to confirm the graphical approximation I made before it? Or am I looking for another method? (Relating to the 'confirm your results by differentiation' comment in the question)

2) Its asks for the rate of change of current at t=0. I am aware that there is no current through the inductor at t=0 due to the inductor opposing the current change (back emf etc) but there IS a current rate of change at t=0 and it can be calculated as: initial di/dt = E/L Amps. If I use this I get 50A/s as an initial current rate of change. How can this be shown graphically on the growth curve and is the above initial di/dt equation enough to satisfy the 'confirm by differentiation' statement in the question?

Thanks in advance.
Simon

You are probably in the wrong forum here. I think Introductory Physics would be better. The only problems I see is with the units you are getting for the graphical approximation. For example you predict the dI/dt at 10ms to be 30mA/s. I get approximately 30A/s. Can you check how you are getting units like mA/s? I'll bet you've got the time axis labelled in seconds instead of milliseconds.
 
  • #3
Thank you for your responses, maybe a mod could move the question elsewhere if this is not the correct place?

As for answering the question in your reply...

I am an idiot its official! Schoolboy error!

The graph is labelled in ms, but I was taking the reading on the calculator to be mA/s not A/s. No idea why, probably due to the fact that I have been trying to work this out for 4 days now.

That now explains how there is an initial di/dt of 50A but again how to show this graphically? Or is it a case of creating the old right angled triangle and di/dt from the two points again?
 
  • #4
Dick
Science Advisor
Homework Helper
26,263
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Thank you for your responses, maybe a mod could move the question elsewhere if this is not the correct place?

As for answering the question in your reply...

I am an idiot its official! Schoolboy error!

The graph is labelled in ms, but I was taking the reading on the calculator to be mA/s not A/s. No idea why, probably due to the fact that I have been trying to work this out for 4 days now.

That now explains how there is an initial di/dt of 50A but again how to show this graphically? Or is it a case of creating the old right angled triangle and di/dt from the two points again?

Right. The same way as you did before. Pick a point close to the initial point and find the slope of the secant line connecting the two points.
 
  • #5
Excellent, thank you ever so much. I know I was on the right lines, just needed pointing in the right direction.

Just to clarify, is the differentiation I have included enough to cover that part of the question would you think?
 
  • #6
Dick
Science Advisor
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Excellent, thank you ever so much. I know I was on the right lines, just needed pointing in the right direction.

Just to clarify, is the differentiation I have included enough to cover that part of the question would you think?

I think so. Graphing gives you approximate numbers for the rate of change. Differentiation gives you exact ones. You calculated di/dt at 10ms and got about 30A/s, right?
 
  • #7
Yes, calculated di/dt from 10.25ms in increments to 10.001ms and was presented with 30A/s every time so took that to prove that 10ms would equal 30A/s too.
 

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