# Rate Of Change Of Current In An Inductor

Discussion in 'Homework Help' started by Simon_G, Dec 21, 2012.

1. ### Simon_G Thread Starter New Member

Jan 15, 2012
2
0
1. The problem

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.

2. Relevant 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)

3. What I have done so far any my question (s)

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?

Simon

Last edited: Dec 21, 2012
2. ### Simon_G Thread Starter New Member

Jan 15, 2012
2
0
Cant edit the above for some reason, but have to point out that all mA/s values above should read A/s. My fault!

3. ### WBahn Moderator

Mar 31, 2012
18,089
4,917
You have an equation for i(t) that you've given above. Differentiate it with respect to t to get di(t)/dt.

On your graph, you want the tangent line at t=0s, not at t=10ms. So use points at t=0 (which is easy since i(t=0s)=0A) and i(t=T) for T just a bit larger than 0s (say, t=1ms). Show that, as T gets smaller, the answer approaches what you get from direct differentiation.