A Zener diode of power rating 1.6,W is to be | vedclass

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(C) The power rating of the Zener diode is $P = 1.6\,W$ and the breakdown voltage is $V_z = 8\,V$.The maximum current that can flow through the Zener

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$10$

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(C) The power rating of the Zener diode is $P = 1.6\,W$ and the breakdown voltage is $V_z = 8\,V$.The maximum current that can flow through the Zener diode is $I_{z,max} = \frac{P}{V_z} = \frac{1.6\,W}{8\,V} = 0.2\,A$.For the Zener diode to act as a voltage regulator, the input voltage $V_{in}$ must be greater than the breakdown voltage $V_z$. Here, the input voltage fluctuates between $3\,V$ and $10\,V$. The Zener diode will only regulate when $V_{in} \geq 8\,V$. Therefore, we consider the maximum input voltage $V_{in,max} = 10\,V$ to ensure the diode does not exceed its power rating.The voltage drop across the series resistance $R_s$ at maximum input voltage is $V_{R_s} = V_{in,max} - V_z = 10\,V - 8\,V = 2\,V$.To ensure safe operation, the current through the series resistance $R_s$ must be equal to the maximum current the Zener diode can handle (assuming no load current is connected, which is the worst-case scenario for the diode).Thus, $R_s = \frac{V_{R_s}}{I_{z,max}} = \frac{2\,V}{0.2\,A} = 10\,\Omega$.

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