Understanding Self-Induction: Calculating Current Growth Rate

[sagda_m2]

This question is about the self induction:

The resistance of a coil is 15 ohm, itself induction is 0.6 Henry, it is connected to a source of direct current which gives 120 volts. Calculate the rate of growth of the current at the moment when the current reaches 80 % of its maximum value.

My Answer:

First I calculated the maximum current:
I= V/R = 120/5 =8 Ampere

Therefore: the 80 % of the maximum current =6.4 A

Then the electromotive force will be:
V= I * R = 6.4 * 15 = 96 volt

And since: the induced e.m.f in the coil = -L * (dI/dt)

Therefore: 96 = 0.6 * (dI/dt)

Therefore: dI/dt = 96/0.6 = 160 A/s

But…..
I found the answer in my book as follows:

Since e.m.f = -L*(dI/dt)

Therefore: (120-96) = 0.6 * (dI/dt)

Therefore: dI/dt = 24/0.6 = 40 A/s

So, why did he substitute the e.m.f in the equation with the difference in volts (120-96)? Why didn't he substitute with the (96 volts)?


Thanks in advance,

 

[chrisk]

Consider the emf produced by the coil and the battery as separate and think about Lenz' Law.

 

[nlightner]

I would say that both approaches are correct and valid ways to calculate the rate of growth of the current in this scenario. The difference in approach may be due to different interpretations of the problem or different assumptions made. In the first approach, the e.m.f is calculated based on the 80% of the maximum current, while in the second approach, it is calculated based on the difference in voltage between the source and the induced e.m.f. Both methods can give accurate results, but it is important to clearly state the assumptions and interpretations made in the calculation. It is also worth considering the context in which the problem is presented and any specific instructions or guidelines given. Ultimately, the important thing is to understand the concept of self-induction and how to calculate the rate of current growth in such scenarios.