A voltage regulating circuit consisting of Zener diode having breakdown voltage of 10 V and maximum power dissipation of 0.4 W is operated at 15 V
Q. A voltage regulating circuit consisting of Zener diode, having break-down voltage of 10 V and maximum power dissipation of 0.4 W, is operated at 15 V. The approximate value of protective resistance in this circuit is ____ Ω.

Correct Answer: 125 Ω

Explanation (Complete Step by Step Calculation)

In a Zener voltage regulator, the protective resistance limits the current through the Zener diode so that its power rating is not exceeded.

Given Zener breakdown voltage:

\[ V_Z = 10 \, \text{V} \]

Maximum power dissipation of Zener diode:

\[ P_{max} = 0.4 \, \text{W} \]

Maximum Zener current is:

\[ I_{Z(max)} = \frac{P_{max}}{V_Z} \]
\[ I_{Z(max)} = \frac{0.4}{10} = 0.04 \, \text{A} \]

Supply voltage:

\[ V_S = 15 \, \text{V} \]

Voltage across the protective resistance is:

\[ V_R = V_S - V_Z = 15 - 10 = 5 \, \text{V} \]

Protective resistance is:

\[ R = \frac{V_R}{I_{Z(max)}} \]
\[ R = \frac{5}{0.04} = 125 \, \Omega \]

Hence, the approximate value of protective resistance is 125 Ω.

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