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.