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Consider the shunt regulator circuit shown infigure withRs. = 0.75 kΩ. Rl = 1KΩ, Vs= 15Vzener breakdown voltage (Vz) = 9V, rated power of zener diodePzen = 25 mW, thena)The load voltage (VL) is 9V.b)The load voltage (VL) is 8.57Vc)If Rc is changed to 1.25 kΩ then power dissipated in zener diode is 2.7mW.d)If Rl is changed to 1.25 kΩthen power dissipated in zener diode is 7.2mWCorrect answer is option 'B,D'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Consider the shunt regulator circuit shown infigure withRs. = 0.75 kΩ. Rl = 1KΩ, Vs= 15Vzener breakdown voltage (Vz) = 9V, rated power of zener diodePzen = 25 mW, thena)The load voltage (VL) is 9V.b)The load voltage (VL) is 8.57Vc)If Rc is changed to 1.25 kΩ then power dissipated in zener diode is 2.7mW.d)If Rl is changed to 1.25 kΩthen power dissipated in zener diode is 7.2mWCorrect answer is option 'B,D'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the shunt regulator circuit shown infigure withRs. = 0.75 kΩ. Rl = 1KΩ, Vs= 15Vzener breakdown voltage (Vz) = 9V, rated power of zener diodePzen = 25 mW, thena)The load voltage (VL) is 9V.b)The load voltage (VL) is 8.57Vc)If Rc is changed to 1.25 kΩ then power dissipated in zener diode is 2.7mW.d)If Rl is changed to 1.25 kΩthen power dissipated in zener diode is 7.2mWCorrect answer is option 'B,D'. Can you explain this answer?.

Solutions for Consider the shunt regulator circuit shown infigure withRs. = 0.75 kΩ. Rl = 1KΩ, Vs= 15Vzener breakdown voltage (Vz) = 9V, rated power of zener diodePzen = 25 mW, thena)The load voltage (VL) is 9V.b)The load voltage (VL) is 8.57Vc)If Rc is changed to 1.25 kΩ then power dissipated in zener diode is 2.7mW.d)If Rl is changed to 1.25 kΩthen power dissipated in zener diode is 7.2mWCorrect answer is option 'B,D'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE. Download more important topics, notes, lectures and mock test series for GATE Exam by signing up for free.

Here you can find the meaning of Consider the shunt regulator circuit shown infigure withRs. = 0.75 kΩ. Rl = 1KΩ, Vs= 15Vzener breakdown voltage (Vz) = 9V, rated power of zener diodePzen = 25 mW, thena)The load voltage (VL) is 9V.b)The load voltage (VL) is 8.57Vc)If Rc is changed to 1.25 kΩ then power dissipated in zener diode is 2.7mW.d)If Rl is changed to 1.25 kΩthen power dissipated in zener diode is 7.2mWCorrect answer is option 'B,D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Consider the shunt regulator circuit shown infigure withRs. = 0.75 kΩ. Rl = 1KΩ, Vs= 15Vzener breakdown voltage (Vz) = 9V, rated power of zener diodePzen = 25 mW, thena)The load voltage (VL) is 9V.b)The load voltage (VL) is 8.57Vc)If Rc is changed to 1.25 kΩ then power dissipated in zener diode is 2.7mW.d)If Rl is changed to 1.25 kΩthen power dissipated in zener diode is 7.2mWCorrect answer is option 'B,D'. Can you explain this answer?, a detailed solution for Consider the shunt regulator circuit shown infigure withRs. = 0.75 kΩ. Rl = 1KΩ, Vs= 15Vzener breakdown voltage (Vz) = 9V, rated power of zener diodePzen = 25 mW, thena)The load voltage (VL) is 9V.b)The load voltage (VL) is 8.57Vc)If Rc is changed to 1.25 kΩ then power dissipated in zener diode is 2.7mW.d)If Rl is changed to 1.25 kΩthen power dissipated in zener diode is 7.2mWCorrect answer is option 'B,D'. Can you explain this answer? has been provided alongside types of Consider the shunt regulator circuit shown infigure withRs. = 0.75 kΩ. Rl = 1KΩ, Vs= 15Vzener breakdown voltage (Vz) = 9V, rated power of zener diodePzen = 25 mW, thena)The load voltage (VL) is 9V.b)The load voltage (VL) is 8.57Vc)If Rc is changed to 1.25 kΩ then power dissipated in zener diode is 2.7mW.d)If Rl is changed to 1.25 kΩthen power dissipated in zener diode is 7.2mWCorrect answer is option 'B,D'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Consider the shunt regulator circuit shown infigure withRs. = 0.75 kΩ. Rl = 1KΩ, Vs= 15Vzener breakdown voltage (Vz) = 9V, rated power of zener diodePzen = 25 mW, thena)The load voltage (VL) is 9V.b)The load voltage (VL) is 8.57Vc)If Rc is changed to 1.25 kΩ then power dissipated in zener diode is 2.7mW.d)If Rl is changed to 1.25 kΩthen power dissipated in zener diode is 7.2mWCorrect answer is option 'B,D'. Can you explain this answer? tests, examples and also practice GATE tests.