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JEE Main 201910 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Evening Shift
Subject
Physics
Chapter
Semiconductors

JEE Main 2019 · 10 April · Evening Shift · Physics · Semiconductors

The figure represents a voltage regulator circuit using a Zener...

The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6V6V and the load resistance is, RL=4kΩ{R}_{L} = 4kΩ . The series resistance of the circuit is Ri=1kΩ{R}_{i} = 1kΩ . If the battery voltage VB{V}_{B} varies from 8V8V to 16V16V , what are the minimum and maximum values of the current through Zener diode?
The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6V

Options

  1. A0.5mA;6mA0.5mA; 6mA
  2. B1mA;8.5mA1mA; 8.5mA
  3. C0.5mA;8.5mA0.5mA; 8.5mA
  4. D1.5mA;8.5mA1.5mA; 8.5mA

Correct answer

C. 0.5mA;8.5mA0.5mA; 8.5mA

Step-by-step solution

For V0=8V{V}_{0} = 8V
The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6V
8V−1000i0=6V8V - 1000{i}_{0} = 6V
i0=2mA{i}_{0} = 2mA
i0−i1=64000=1.5mA{i}_{0} - {i}_{1} = \dfrac{6}{4000} = 1.5mA
i1=0.5mA{i}_{1} = 0.5mA
For V0=16V{V}_{0} = 16V
The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6V
16V−1000i0=6V16V - 1000{i}_{0} = 6V
i0=10mA{i}_{0} = 10mA
i0−i2=64000=1.5mA{i}_{0} - {i}_{2} = \dfrac{6}{4000} = 1.5mA
i2=8.5mA{i}_{2} = 8.5mA
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