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(A) Given:Zener voltage $V_z = 5.6 \, V$Maximum power dissipation $P_{z \max} = \frac{1}{4} \, W = 0.25 \, W$Input voltage $V_{in} = 10 \, V$The maxim

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$98.56$

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(A) Given:Zener voltage $V_z = 5.6 \, V$Maximum power dissipation $P_{z \max} = \frac{1}{4} \, W = 0.25 \, W$Input voltage $V_{in} = 10 \, V$The maximum current $I_z$ that the zener diode can handle is given by:$P_{z \max} = V_z 	imes I_z$$0.25 = 5.6 	imes I_z$$I_z = \frac{0.25}{5.6} \, A$In the circuit, the current through the series resistance $R_s$ is $I_s = I_z$. The voltage drop across the resistance $R_s$ is:$V_{R_s} = V_{in} - V_z = 10 \, V - 5.6 \, V = 4.4 \, V$Using Ohm's law for the resistance $R_s$:$R_s = \frac{V_{R_s}}{I_s} = \frac{4.4}{I_z} = \frac{4.4}{(0.25 / 5.6)}$$R_s = \frac{4.4 	imes 5.6}{0.25} = 4.4 	imes 5.6 	imes 4 = 98.56 \, \Omega$Thus, the minimum value of the resistance $R_s$ is $98.56 \, \Omega$.

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