Semiconductors

JEE Main level, numerical answer, medium

34 of 209

Question 34 of 209

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.

Show answer and solution

Answer: 125

The maximum current through the Zener diode is determined by its maximum power dissipation: IZ,max=PmaxVZ=0.410=0.04I_{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

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Semiconductors, diodes and logic gates simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Semiconductors (Diodes), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.