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A Zener diode connected across a source voltage of 12V such that 6V drops across it. The power drawn by zener diode is 2.4 MW, Then what is the maximum value of series resistance connected in the circuit?
Correct answer is '15'. Can you explain this answer?
Verified Answer
A Zener diode connected across a source voltage of 12V such that 6V dr...
VS = 12 Volt
Vdrop = 6 Volt
So, remaining voltage V = VS – Vdrop = 12 – 6
= 6 Volt
Power drawn by Zener diode P = 2.4 MW
As we know that
Vdrop = 6 Volt
So, remaining voltage V = VS – Vdrop = 12 – 6
= 6 Volt
Power drawn by Zener diode P = 2.4 MW
As we know that
Most Upvoted Answer
A Zener diode connected across a source voltage of 12V such that 6V dr...
To find the maximum value of the series resistance connected in the circuit, we need to analyze the given information step by step.
Step 1: Determine the voltage across the series resistance
Since the Zener diode has a voltage drop of 6V across it, the remaining voltage across the series resistance can be calculated as follows:
Voltage across the series resistance = Source voltage - Zener diode voltage
Voltage across the series resistance = 12V - 6V = 6V
Step 2: Calculate the power dissipated by the series resistance
To calculate the power dissipated by the series resistance, we can use the formula:
Power = (Voltage)^2 / Resistance
Given that the power drawn by the Zener diode is 2.4 MW (megawatts), we can convert it to watts by multiplying by 10^6:
Power = 2.4 MW = 2.4 x 10^6 W
Substituting the values into the formula, we have:
2.4 x 10^6 W = (6V)^2 / Resistance
Step 3: Solve for the resistance
Rearranging the formula, we have:
Resistance = (6V)^2 / (2.4 x 10^6 W)
Simplifying the expression, we get:
Resistance = 36V^2 / (2.4 x 10^6 W)
Resistance = 36 / (2.4 x 10^6) Ohms
Resistance = 15 Ohms
Therefore, the maximum value of the series resistance connected in the circuit is 15 Ohms.
In summary,
- The voltage across the series resistance is 6V.
- The power dissipated by the series resistance is 2.4 MW.
- By using the formula for power dissipation, we can calculate the resistance to be 15 Ohms.
Step 1: Determine the voltage across the series resistance
Since the Zener diode has a voltage drop of 6V across it, the remaining voltage across the series resistance can be calculated as follows:
Voltage across the series resistance = Source voltage - Zener diode voltage
Voltage across the series resistance = 12V - 6V = 6V
Step 2: Calculate the power dissipated by the series resistance
To calculate the power dissipated by the series resistance, we can use the formula:
Power = (Voltage)^2 / Resistance
Given that the power drawn by the Zener diode is 2.4 MW (megawatts), we can convert it to watts by multiplying by 10^6:
Power = 2.4 MW = 2.4 x 10^6 W
Substituting the values into the formula, we have:
2.4 x 10^6 W = (6V)^2 / Resistance
Step 3: Solve for the resistance
Rearranging the formula, we have:
Resistance = (6V)^2 / (2.4 x 10^6 W)
Simplifying the expression, we get:
Resistance = 36V^2 / (2.4 x 10^6 W)
Resistance = 36 / (2.4 x 10^6) Ohms
Resistance = 15 Ohms
Therefore, the maximum value of the series resistance connected in the circuit is 15 Ohms.
In summary,
- The voltage across the series resistance is 6V.
- The power dissipated by the series resistance is 2.4 MW.
- By using the formula for power dissipation, we can calculate the resistance to be 15 Ohms.
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A Zener diode connected across a source voltage of 12V such that 6V drops across it. The power drawn by zener diode is 2.4 MW, Then what is the maximum value of series resistance connected in the circuit?Correct answer is '15'. Can you explain this answer?
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A Zener diode connected across a source voltage of 12V such that 6V drops across it. The power drawn by zener diode is 2.4 MW, Then what is the maximum value of series resistance connected in the circuit?Correct answer is '15'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A Zener diode connected across a source voltage of 12V such that 6V drops across it. The power drawn by zener diode is 2.4 MW, Then what is the maximum value of series resistance connected in the circuit?Correct answer is '15'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A Zener diode connected across a source voltage of 12V such that 6V drops across it. The power drawn by zener diode is 2.4 MW, Then what is the maximum value of series resistance connected in the circuit?Correct answer is '15'. Can you explain this answer?.
A Zener diode connected across a source voltage of 12V such that 6V drops across it. The power drawn by zener diode is 2.4 MW, Then what is the maximum value of series resistance connected in the circuit?Correct answer is '15'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A Zener diode connected across a source voltage of 12V such that 6V drops across it. The power drawn by zener diode is 2.4 MW, Then what is the maximum value of series resistance connected in the circuit?Correct answer is '15'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A Zener diode connected across a source voltage of 12V such that 6V drops across it. The power drawn by zener diode is 2.4 MW, Then what is the maximum value of series resistance connected in the circuit?Correct answer is '15'. Can you explain this answer?.
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