An ideal diode is shown in the above figure. If `V_(i)` changes from `1V to 8V`, then find the change in the current in Milliamperes:

Answer

`25.00`

Explanation

When Vi = IV, the diode is reverse biased.
`therefore I_("initial") = 0`
When `V_(i) = 8V, V_("final") = V_(i) -V = 8-4 = 4V`
`therefore I_("final") = 4/160 = 0.025A`
`therefore` Change in I = 0.025 A = 25 mA