# Why is the derivate used in the faraday equation?

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.

• 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. – C. Towne Springer Jun 1 '17 at 5:56