A zener diode with 5V zener voltage is used to regulate an unregulated dc voltage input of 25 V. For a 400 $$\Omega$$ resistor connected in series, the zener current is found to be 4 times load current. The load current ($$I_L$$) and load resistance ($$R_L$$) are:

The zener diode keeps the voltage across itself and the parallel load fixed at $$V_Z = 5\text{ V}$$.

The source voltage is $$V_S = 25\text{ V}$$ and the series resistor is $$R_S = 400\,\Omega$$.

Hence the voltage across $$R_S$$ equals $$V_S - V_Z = 25 - 5 = 20\text{ V}$$.

Current through the series resistor (which is the sum of zener current $$I_Z$$ and load current $$I_L$$) is therefore
$$I_S = \frac{20}{400} = 0.05\text{ A} = 50\text{ mA}$$ $$-(1)$$

Given that the zener current is four times the load current,
$$I_Z = 4 I_L$$ $$-(2)$$

From $$I_S = I_Z + I_L$$ and substituting from $$(2)$$:
$$I_S = 4 I_L + I_L = 5 I_L$$ $$-(3)$$

Using $$(1)$$ and $$(3)$$:
$$5 I_L = 50\text{ mA} \;\Longrightarrow\; I_L = 10\text{ mA}$$

The load resistance is found from Ohm’s law using the regulated voltage:
$$R_L = \frac{V_Z}{I_L} = \frac{5\text{ V}}{10\text{ mA}} = 500\,\Omega$$

Thus, $$I_L = 10\text{ mA}$$ and $$R_L = 500\,\Omega$$. This matches Option D.