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The value of power dissipated across the Zener diode ....

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 $

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