A voltage regulating circuit consisting of a 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 ______ . [2026]

Question

A voltage regulating circuit consisting of a 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 ______ $Ω$ . [2026]

Smart Achievers · Accepted Answer

(125)

$P_{D} = 0.4 W = 10 i$

$i = 0.04 A$

$R = \frac{15 - 10}{0.04} = \frac{5}{0.04} = 125 Ω$

Solution

Q.
A voltage regulating circuit consisting of a 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 ______ $Ω$ . [2026]

Ans.
(125)

$P_{D} = 0.4 W = 10 i$

$i = 0.04 A$

$R = \frac{15 - 10}{0.04} = \frac{5}{0.04} = 125 Ω$

Want Us to Email you About Special Offers & Updates?

Solution

Q. A voltage regulating circuit consisting of a 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. [2026]

Ans. (125) PD=0.4 W=10i i=0.04 A R=15-100.04=50.04=125 Ω

Want Us to Email you About Special Offers & Updates?

Q.
A voltage regulating circuit consisting of a 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 ______ $Ω$ . [2026]

Ans.
(125)

$P_{D} = 0.4 W = 10 i$

$i = 0.04 A$

$R = \frac{15 - 10}{0.04} = \frac{5}{0.04} = 125 Ω$