In the figure shown, the maximum and minimum currents through Zener diode are

(l) 5 mA, l mA               (2) 14 mA, 6 mA                    (3) 14 mA, 5 rnA                   (4) 9 mA, 1 mA

Solution

For a zener diode operating in the breakdown region as a voltage regulator, the current through the input resistance R_S is given by,

$\mathrm{I} = \frac{{\mathrm{V}}_{\mathrm{S}} - {\mathrm{V}}_{\mathrm{z}}}{{\mathrm{R}}_{\mathrm{S}}} \left(1\right)$    where V_S is the supply voltage and V_z is the voltage across zener diode
The​ current through the zener diode can be expressed as ,

I_Z = I - I_L       (2), where I_L is the load current through the variable load resistance R_L and is given by I_L =$\frac{{\mathrm{V}}_{\mathrm{O}}}{{\mathrm{R}}_{\mathrm{L}}} =\frac{{\mathrm{V}}_{\mathrm{Z}}}{{\mathrm{R}}_{\mathrm{L}}}$  (3)

We have, V_Smax = 120V, V_Smin = 80V, V_Z = 50V , R_S = 5 k$\Omega$ and R_L = 10 k​​$\Omega$
Then I_L = 5 mA                     (using equation (3))

I_max = $\frac{{\mathrm{V}}_{\mathrm{Smax}} - {\mathrm{V}}_{\mathrm{Z}}}{{\mathrm{R}}_{\mathrm{S}}} = \frac{120-50}{5\times {10}^{-3}} =$ 14 mA;  Then I_Zmax = 9 mA     (using equation (2))
and

${\mathrm{I}}_{\min }=\frac{{\mathrm{V}}_{\mathrm{Smin}} - {\mathrm{V}}_{\mathrm{Z}}}{{\mathrm{R}}_{\mathrm{S}}} = \frac{80-50}{5\times {10}^{-3}}$  = 6 mA ;   Then I_Zmin  = 1 mA

Thus option (4) is the right choice