Concept:The series current-limiting resistor \( R_s \) must drop the excess voltage (\( V_{\text{in}} - V_Z \)) while supplying the combined current for both the load (\( I_L \)) and the Zener diode (\( I_Z \)).
Formula:$$I_{\text{total}} = I_Z + I_L$$
$$R_s = \frac{V_{\text{in}} - V_Z}{I_{\text{total}}} = \frac{V_{\text{in}} - V_Z}{I_Z + I_L}$$
Solution:- Voltage drop across the series resistor: \( V_R = V_{\text{in}} - V_Z = 12.0\text{ V} - 6.0\text{ V} = 6.0\text{ V} \).
- Total circuit current: \( I_{\text{total}} = I_Z + I_L = 5.0\text{ mA} + 20\text{ mA} = 25\text{ mA} = 0.025\text{ A} \).
- Required resistance: \( R_s = \frac{6.0\text{ V}}{0.025\text{ A}} = 240\ \Omega \).
Why other options are incorrect:- Option A: 120 Ω allows a total current of 50 mA, causing unnecessary power dissipation.
- Option C: 300 Ω limits total current to 20 mA, starving the Zener diode of its minimum operating current.
- Option D: 600 Ω limits current to only 10 mA, causing the output voltage to drop out of regulation.
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