A semiconductor diode and a resistor of constant resistance are connected in some way inside a box having two external terminals. When a potential difference `V` of `1V` is applied, `1=25 mA`. If potential difference is reversed, `I=50 mA`. Forward resistance and diode resistance are

Options

A. `40 Omega` and `40 Omega`
B. `0 Omega , oo`
C. `Omega , 12 Omega`
D. `40 Omega , 20 Omega `

Answer

A

Explanation

The value of R remains constant. But D conducts when it is forward biased but when it is reverse biased it does not conduct.
In the given problem, the current flows in both the cases i.e. D and R are not in seri es otherwise the current would have been zero in one case. Hence D and R are joined in parallel.
I = 25 mA corresponds to rever-se biasing of R and D In that case, the current flows only through R.
`therefore R=V/I=1/(25xx10^(-3)) = 40 Omega`
But when D is forward biased, it has a small resistance and as D and R are in parallel, their combined resistance, `R_P=(1V)/(50xx10^(-3)) = 20 Omega`
`therefore` Diode resistance will be 40 `Omega [ because 1/20 =1/40 +1/40]`
Thus R= 40 `Omega` and Forward resistance = `40 Omega`