A Zener diode with a power rating of 1 , W is | vedclass

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(B) The power rating of the Zener diode is $P = 1 \, W$ and its breakdown voltage is $V_Z = 5 \, V$.The maximum current that can flow through the Zene

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$10 \, \Omega$

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(B) The power rating of the Zener diode is $P = 1 \, W$ and its breakdown voltage is $V_Z = 5 \, V$.The maximum current that can flow through the Zener diode is given by $I_{Z, 	ext{max}} = \frac{P}{V_Z} = \frac{1 \, W}{5 \, V} = 0.2 \, A$.For the circuit to operate safely,the Zener diode must handle the maximum input voltage without exceeding its power rating. The voltage across the series resistor $R_s$ is $V_{R_s} = V_i - V_Z$.To ensure the Zener diode is not damaged,we consider the maximum input voltage $V_{i, 	ext{max}} = 7 \, V$.The current through the resistor $R_s$ is $I = \frac{V_{i, 	ext{max}} - V_Z}{R_s}$.For safe operation,this current must be at least the maximum current the Zener can handle: $I \ge I_{Z, 	ext{max}}$.Thus,$R_s \le \frac{V_{i, 	ext{max}} - V_Z}{I_{Z, 	ext{max}}} = \frac{7 \, V - 5 \, V}{0.2 \, A} = \frac{2 \, V}{0.2 \, A} = 10 \, \Omega$.Therefore,the value of $R_s$ should be $10 \, \Omega$.

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