Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main 202624 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Morning Shift
Subject
Physics
Chapter
Semiconductors

JEE Main 2026 · 24 January · Morning Shift · Physics · Semiconductors

A voltage regulating circuit consisting of Zener diode, having...

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\Omega.

Correct answer

125

Step-by-step solution

The maximum current through the Zener diode is determined by its maximum power dissipation: I_{Z,max} = \frac{P_\max}{V_Z} = \frac{0.4}{10} = 0.04 A.
The protective series resistance must limit the current such that the Zener operates safely. Using Ohm's law across the series resistance and Zener when the Zener carries maximum current:
Rs=Vin−VZIZ,max=15−100.04=50.04=125R_s = \frac{V_{in} - V_Z}{I_{Z,max}} = \frac{15 - 10}{0.04} = \frac{5}{0.04} = 125 Ω
Practice Semiconductors on Quantrex Academy →

More from Semiconductors

The output Y for the given inputs A and B to the circuit is: JEE Main 2026Refer to the logic circuit given below. For two inputs (A=1, B=1) and (A=0, B=1) , output… JEE Main 2026The maximum rated power of the LED is 2 mW and it is used in the circuit with input… JEE Main 2026In a semiconductor p-n diode, the doping concentrations on p-side and n-side are 10^ 15… JEE Main 2026For the given logic circuit, which of the following inputs combination will make both LED… JEE Main 2026Given below are two statements: one is labelled as Assertion A and the other is labelled… JEE Main 2026A diode has Zener voltage of 10 V and maximum power dissipation of 0.5 W, then the… JEE Main 2026Two 4 bits binary numbers, A = 1101 and B = 1010 are given in the inputs of a logic… JEE Main 2026 Full Semiconductors list All JEE Main PYQs