A junction diode has a resistance of 25 `Omega` when forward biased and 2500 `Omega` when reverse biased. What is the current in the diode, for the arrangement shown ?

Options

A. `1/25A`
B. `1/7A`
C. `1/35A`
D. `1/480A`

Answer

B

Explanation

In this case, V = 5 V. The diode is forward biased.
Net resistance= `10Omega +25 Omega = 35 Omega`
`therefore I=V/R =5/35 =1/7A`
(No current in reverse bias)