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

The value is
`\Delta t = 4.0 *10^(-7) \ s`

Step-by-step 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 * (\Delta I )/(\Delta t )`

Here
`\Delta t` is the time taken

=>
`\Delta t = (L * \Delta I )/(\epsilon )`

=>
`\Delta t = (2*10^(-3) * 0.100 )/(500 )`

=>
`\Delta t = 4.0 *10^(-7) \ s`