0
$\begingroup$

Sorry for this silly question, but I need a little bit "intuitive" definition of derivate. For example, we have the faraday law:

$$E =L * \frac{\partial i}{\partial t}$$

I know the more quickly chance in current $\partial i$ in the same time $\partial t$ become more voltage. But I have this example :

Imagine that an inductor of 200mH connected across a supply of 9V is passing a current of 2amperes. When the current is switched off, it collapses to zero in 10ms, what would be the back emf generated across the coil?

E = 200mH x 2A / 10ms

or

E =200 x 10-3 x 2/10 x 10-3

= 40volts

So the back emf generated at switch off is more than 4 times higher than the supply voltage!

So my question is: how is the derivate used there, because in the example i just see integers numbers.

Best regards.

$\endgroup$
  • $\begingroup$ The current goes from 2A to zero in 10ms. The problem assumes the current decreases linearly. This means the derivative with respect to time is the same as the change in current over the interval of 10ms. They are using delta A/delta t. $\endgroup$ – C. Towne Springer Jun 1 '17 at 5:56
0
$\begingroup$

So basically a derivative can be seen as dy/dx or Δy/Δx. Here Δy/Δx is Δi/Δt, which is equal to 200 amp/sec. Hence the back emf is 40 Volts.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.