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(A) The voltage drop across the series resistor $R_s$ is given by:$V_s = V_{in} - V_z = 8 \, V - 5 \, V = 3 \, V$The current through the load resistor

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$0.6 \, k \Omega$

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(A) The voltage drop across the series resistor $R_s$ is given by:$V_s = V_{in} - V_z = 8 \, V - 5 \, V = 3 \, V$The current through the load resistor $R_L$ is:$I_L = \frac{V_z}{R_L} = \frac{5 \, V}{1 	imes 10^3 \, \Omega} = 5 \, mA$The maximum current through the Zener diode is determined by its power rating $P_z = 10 \, mW$:$I_{z,max} = \frac{P_z}{V_z} = \frac{10 \, mW}{5 \, V} = 2 \, mA$For the Zener diode to regulate, the total current $I_s$ must be $I_L + I_z$.The Zener current must be between $0$ and $I_{z,max}$.Thus, $I_{s,min} = I_L + 0 = 5 \, mA$ and $I_{s,max} = I_L + I_{z,max} = 5 \, mA + 2 \, mA = 7 \, mA$.The resistance $R_s$ must satisfy:$R_{s,min} = \frac{V_s}{I_{s,max}} = \frac{3 \, V}{7 \, mA} = \frac{3}{7} \, k \Omega \approx 0.43 \, k \Omega$$R_{s,max} = \frac{V_s}{I_{s,min}} = \frac{3 \, V}{5 \, mA} = 0.6 \, k \Omega$Therefore, the value of $R_s$ must be in the range $0.43 \, k \Omega \le R_s \le 0.6 \, k \Omega$.Among the given options, $0.6 \, k \Omega$ is the only valid value.

In the given circuit, if the power rating of the Zener diode is $10 \, mW$, the value of series resistance $R_s$ to regulate the input unregulated supply is:

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