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(B) The load resistance $R_L$ is given by $R_L = \frac{V_L}{I_L} = \frac{110 \, V}{250 	imes 10^{-3} \, A} = 440 \, \Omega$.Given that the maximum cu

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$R_L = 440 \, \Omega, R_S = 140 \, \Omega$

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(B) The load resistance $R_L$ is given by $R_L = \frac{V_L}{I_L} = \frac{110 \, V}{250 	imes 10^{-3} \, A} = 440 \, \Omega$.Given that the maximum current $I$ is equally shared between the load and the Zener diode, we have $I_Z = I_L = 250 \, mA$.Therefore, the total maximum current is $I = I_L + I_Z = 250 \, mA + 250 \, mA = 500 \, mA = 0.5 \, A$.The series resistance $R_S$ is calculated at the maximum input voltage $V_{in,max} = 180 \, V$ as $R_S = \frac{V_{in,max} - V_L}{I} = \frac{180 \, V - 110 \, V}{0.5 \, A} = \frac{70 \, V}{0.5 \, A} = 140 \, \Omega$.Thus, $R_L = 440 \, \Omega$ and $R_S = 140 \, \Omega$.

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