The value of power dissipated across the Zener diode ($V_Z = 15 V $) connected in the circuit as shown in the figure is $x \times 10^{-1} Watt$. The value of x, to the nearest integer, is _ _ _ _ .

Solution

$22 V = V_{R_S} + V_Z $

$\therefore V_{R_S} = 22 – 15 = 7 V $

Current supplied by 22 V = $I = \frac {7}{35} = \frac {1}{5} A$

$I_{R_L} = \frac {15}{90} = \frac {1}{6} A$

Now, $I_Z = \frac {1}{5} – \frac {1}{6} = \frac {1}{30} A$

$P_Z = I_Z \times V_Z = \frac {1}{30} \times 15 = 5 \times 10^{-1} Watt$

$\therefore x = 5 $