In a Zener diode,the reverse bias voltage is $3 \ V$ and the width of the depletion region is $300 \ \mathring{A}$. The electric field intensity will be $\dots \ V/cm$.

Two Zener diodes ($A$ and $B$) having breakdown voltages of $6\, V$ and $4\, V$ respectively,are connected as shown in the circuit below. The output voltage $V_{0}$ variation with input voltage linearly increasing with time,is given by: $(V_{\text{input}} = 0\, V$ at $t = 0)$ (figures are qualitative).

The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is $6\,V$ and the load resistance is $R_L = 4\,k\Omega.$ The series resistance of the circuit is $R_i = 1\,k\Omega.$ If the battery voltage $V_B$ varies from $8\,V$ to $16\,V,$ what are the minimum and maximum values of the current through the Zener diode?

$A$ Zener diode having a Zener voltage of $8\, V$ and a power dissipation rating of $0.5\, W$ is connected across a potential divider circuit as shown in the diagram. The value of the protective resistance $R_p$ is $....\, \Omega$.

In the given circuit,the current through the Zener diode is.....$mA$.

Why is there a sudden increase in current in a Zener diode?