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,20 Omega`
B. `40 Omega` and `40 Omega`
C. `0 Omega,oo`
D. `6 Omega,12 Omega`

Answer

B

Explanation

When a diode is reverse-biased, the diode does not conduct. So, if the resistor and diode are it series, then the current should be zero in one of the two given cases. But this is not the case. So, clearly, the two are connected in parallel. Clearly, `I=-25 mA` corresponds to reverse-biasing
Now `R=(1V)/(25xx10^(-3)A)=1000/25 Omega=40 Omega`
Again, `I=50 mA`
Now, current shall flow through the diode also because diode is forward-baised.
If `R_(p)` is the combined resistance of diode and resistor,
then `R_(p)=1/(50xx10^(-3)) Omega=100/50 Omega=20 Omega`
clearly, it is a parallel combination of `40 Omega` and `40 Omega`.