In the following circuit,the current flowing through the Zener diode is (in $\text{ mA}$)

$A$ reverse biased Zener diode when operated in the breakdown region works as

$A$ Zener diode of breakdown voltage $V_{Z} = 8\,V$ and maximum Zener current $I_{ZM} = 10\,mA$ is subjected to an input voltage $V_{i} = 10\,V$ with a series resistance $R = 100\,\Omega$. In the given circuit,$R_{L}$ represents the variable load resistance. The ratio of the maximum and minimum value of $R_{L}$ is

$A$ Zener diode with voltage $V_z = 6 \ V$ is used to maintain a constant voltage across a load resistance $R_L = 1000 \ \Omega$. $A$ series resistor $R_s = 100 \ \Omega$ is used. If the source $e.m.f$ is $E = 9 \ V$,then the power dissipated in the Zener diode is .......... $W$.

$A$ Zener diode of power rating $1.6\,W$ is to be used as a voltage regulator. If the Zener diode has a breakdown voltage of $8\,V$ and it has to regulate an input voltage fluctuating between $3\,V$ and $10\,V$, the value of series resistance $R_s$ for the safe operation of the diode will be $.........\Omega$.

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 $\dots \ V/cm$.