Question

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 $1\, V$ is applied, we get $I=25\, mA$. If potential difference is reversed, $I=50\, mA$. Forward resistance and diode resistance are respectively

• A. $40\, \Omega,\, 20\, \Omega$
• B. $40\, \Omega$ and $40\, \Omega$
• C. $0\, \Omega,\, \infty$
• D. $6\, \Omega,\, 12\, \Omega$

Answer 1

Answer: (B)

When a diode is reverse-biased, the diode does not conduct. So, if the resistor and diode are in 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 $R=\frac{1 V }{25 \times 10^{-3} A }=\frac{1000}{25} \Omega=40\, \Omega$
Again, $I=50\, mA$
Now, current shall flow through the diode also because diode is forward-biased.
If $R_{p}$ is the combined resistance of diode and resistor, then
$R_{p}=\frac{1}{50 \times 10^{-3}}\, \Omega$
$=\frac{1000}{50} \,\Omega=20\, \Omega$
Clearly, it is a parallel combination of $40\, \Omega$ and $40\, \Omega$.