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(D) The input voltage is $V_{in} = 30 \, V$ and the Zener breakdown voltage is $V_Z = 6 \, V$. The load resistance is $R_L = 1 \, k\Omega$.For the Zen

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$4$

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(D) The input voltage is $V_{in} = 30 \, V$ and the Zener breakdown voltage is $V_Z = 6 \, V$. The load resistance is $R_L = 1 \, k\Omega$.For the Zener diode to regulate the voltage,it must operate in the breakdown region. The current through the load resistance is $I_L = \frac{V_Z}{R_L} = \frac{6 \, V}{1 \, k\Omega} = 6 \, mA$.The voltage drop across the series resistor $R_S$ is $V_{R_S} = V_{in} - V_Z = 30 \, V - 6 \, V = 24 \, V$.For the Zener diode to just maintain regulation,the current through the Zener diode $I_Z$ should be at least $0$ (the limiting case for maximum $R_S$).Thus,the total current through $R_S$ is $I_S = I_L + I_Z = 6 \, mA + 0 = 6 \, mA$.Using Ohm's law for $R_S$,we have $R_S = \frac{V_{R_S}}{I_S} = \frac{24 \, V}{6 \, mA} = 4 \, k\Omega$.Therefore,the maximum value of $R_S$ is $4 \, k\Omega$.

In the given circuit,what is the maximum value of $R_S$ (in $k\Omega$) for the Zener diode to regulate the voltage across the load resistance $R_L$?

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