Exams › JEE Advanced › Physics
A Zener diode with breakdown voltage Vz = 9 V and maximum power dissipation Pmax = 0.27 W is used to provide a stabilized 9 V supply from a 12 V battery. The Zener is connected in parallel with the load terminals A and B, and a series resistor Rs is placed between the battery and the Zener/load combination. What is the minimum value of Rs that keeps the Zener within its power rating when no load is connected?
- 111 ohm
- 103 ohm
- 100 ohm
- 97 ohm
Correct answer: 100 ohm
Solution
With no load, the entire current through Rs flows through the Zener. To keep power within Pmax = 0.27 W at Vz = 9 V, the maximum Zener current is 30 mA. The voltage across Rs is 12 - 9 = 3 V, giving Rs_min = 3 / 0.03 = 100 ohm.
Related JEE Advanced Physics questions
⚔️ Practice JEE Advanced Physics free + battle 1v1 →