In the diagram shown, the resistance between points $A$ and $B$ is '$R_1$' when an ideal diode $D$ is forward biased and is '$R_2$' when ideal diode $D$ is reverse biased. The ratio $R_1/R_2$ is

$A$ silicon diode has a forward voltage drop of $1.2\, V$ for a forward $DC$ current of $100\, mA$. It has a reverse current of $1\times10^{-6}\, A$ for a reverse voltage of $10\, V$. The bulk and reverse resistances of the diode are:

The given circuit has two ideal diodes $D_1$ and $D_2$ connected as shown in the figure. The current flowing through the resistance $R_1$ will be (in $\,A$)

In which figure,the junction diode is forward biased?

In the circuit, all three diodes $D_1, D_2, D_3$ have a forward resistance of $50 \text{ } \Omega$ each and infinite backward resistance. If the battery voltage is $5 \text{ V}$, find the current through the $100 \text{ } \Omega$ resistance. (in $\text{mA}$)

Two identical $p-n$ junctions are connected in series with a battery in three ways,as shown below. The potential drops across the two $p-n$ junctions are equal in: