Question

Camera flashes charge a capacitor to high voltage by switching the current through an inductor on and off rapidly. In what time must the 0.100 A current through a 2.00 mH inductor be switched on or off to induce a 500 V emf

Answer 1

Answer:

The value is  $\Delta t = 4.0 *10^{-7} \ s$

Explanation:

From the question we are told that

  The current is $\Delta I = 0.100 \ A$

  The  inductor  is $L = 2.0mH = 2.0*10^{-3} \ H$

  The voltage induced is  $\epsilon = 500 V$

Generally the emf induced is mathematically represented as

      $\epsilon = L * \frac{\Delta I }{\Delta t }$

Here  $\Delta t$ is the time taken  

=>  $\Delta t = \frac{L * \Delta I }{\epsilon }$

=>  $\Delta t = \frac{2*10^{-3} * 0.100 }{500 }$

=>  $\Delta t = 4.0 *10^{-7} \ s$