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A Zener diode has breakdown voltage of 9.1 V with maximum power dissipation of 364m W. What is the maximum current (in amperes) that the diode can with stand if the total output being supplied to the circuit is 910 mW.
Correct answer is '0.04'. Can you explain this answer?
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Understanding Zener Diode Specifications
A Zener diode operates in reverse bias and is characterized by its breakdown voltage and maximum power dissipation. In this case:
- Breakdown Voltage: 9.1 V
- Maximum Power Dissipation: 364 mW
Calculating Maximum Current
To find the maximum current the Zener diode can withstand, we use the formula for power:
- Power (P) = Voltage (V) x Current (I)
Rearranging the formula to find the current gives us:
- Current (I) = Power (P) / Voltage (V)
Substituting the values:
- Maximum Power (P) = 364 mW = 0.364 W
- Breakdown Voltage (V) = 9.1 V
Now, we calculate the maximum current:
- I = 0.364 W / 9.1 V
After performing the calculation, we find:
- I ≈ 0.04 A (or 40 mA)
Comparing with Total Output
The total output supplied to the circuit is 910 mW, which translates to:
- Total Output Power = 0.910 W
This value is greater than the power dissipation capability of the Zener diode (364 mW). Therefore, the Zener should ideally operate below this value to prevent damage.
Conclusion
The Zener diode can withstand a maximum current of 0.04 A based on the power dissipation limit. Operating above this current may lead to overheating and potential failure. Always ensure that the current remains within safe limits when designing circuits with Zener diodes.
A Zener diode operates in reverse bias and is characterized by its breakdown voltage and maximum power dissipation. In this case:
- Breakdown Voltage: 9.1 V
- Maximum Power Dissipation: 364 mW
Calculating Maximum Current
To find the maximum current the Zener diode can withstand, we use the formula for power:
- Power (P) = Voltage (V) x Current (I)
Rearranging the formula to find the current gives us:
- Current (I) = Power (P) / Voltage (V)
Substituting the values:
- Maximum Power (P) = 364 mW = 0.364 W
- Breakdown Voltage (V) = 9.1 V
Now, we calculate the maximum current:
- I = 0.364 W / 9.1 V
After performing the calculation, we find:
- I ≈ 0.04 A (or 40 mA)
Comparing with Total Output
The total output supplied to the circuit is 910 mW, which translates to:
- Total Output Power = 0.910 W
This value is greater than the power dissipation capability of the Zener diode (364 mW). Therefore, the Zener should ideally operate below this value to prevent damage.
Conclusion
The Zener diode can withstand a maximum current of 0.04 A based on the power dissipation limit. Operating above this current may lead to overheating and potential failure. Always ensure that the current remains within safe limits when designing circuits with Zener diodes.
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A Zener diode has breakdown voltage of 9.1 V with maximum power dissipation of 364m W. What is the maximum current (in amperes) that the diode can with stand if the total output being supplied to the circuit is 910 mW.Correct answer is '0.04'. Can you explain this answer? for Physics 2025 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A Zener diode has breakdown voltage of 9.1 V with maximum power dissipation of 364m W. What is the maximum current (in amperes) that the diode can with stand if the total output being supplied to the circuit is 910 mW.Correct answer is '0.04'. Can you explain this answer? covers all topics & solutions for Physics 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A Zener diode has breakdown voltage of 9.1 V with maximum power dissipation of 364m W. What is the maximum current (in amperes) that the diode can with stand if the total output being supplied to the circuit is 910 mW.Correct answer is '0.04'. Can you explain this answer?.
A Zener diode has breakdown voltage of 9.1 V with maximum power dissipation of 364m W. What is the maximum current (in amperes) that the diode can with stand if the total output being supplied to the circuit is 910 mW.Correct answer is '0.04'. Can you explain this answer? for Physics 2025 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A Zener diode has breakdown voltage of 9.1 V with maximum power dissipation of 364m W. What is the maximum current (in amperes) that the diode can with stand if the total output being supplied to the circuit is 910 mW.Correct answer is '0.04'. Can you explain this answer? covers all topics & solutions for Physics 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A Zener diode has breakdown voltage of 9.1 V with maximum power dissipation of 364m W. What is the maximum current (in amperes) that the diode can with stand if the total output being supplied to the circuit is 910 mW.Correct answer is '0.04'. Can you explain this answer?.
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