Which of the following calculates power dissipated by a series linear voltage regulator?
The correct answer is C: Voltage difference from input to output multiplied by output current. Which of the following calculates power dissipated by a series linear voltage regulator is voltage difference from input to output multiplied by output current. Power = (Vin - Vout) × Iout. For amateur radio operators, this is important for power supply design. Understanding this helps when calculating power dissipation.
Exam Tip
Power dissipated by series linear regulator = (Vin - Vout) × Iout. Think 'P'ower 'D'issipated = 'D'ifference 'V'oltage × 'O'utput 'C'urrent. Power = (Vin - Vout) × Iout. Not Vin × Iin, not Vin / Iout, not Vout × Iout - just (Vin - Vout) × Iout.
Memory Aid
"Power dissipated by series linear regulator = (Vin - Vout) × Iout. Think 'P'ower = 'D'ifference × 'C'urrent. Power = (Vin - Vout) × Iout. Important for power supply design."
Real-World Application
Power dissipated by a series linear voltage regulator: It's calculated as voltage difference from input to output multiplied by output current. For example, if Vin=12V, Vout=5V, Iout=1A, then power = (12-5) × 1 = 7W. This is the calculation - (Vin - Vout) × Iout.
Key Concepts
Why Other Options Are Wrong
Option A: Incorrect. Input voltage multiplied by input current isn't correct - power dissipated = (Vin - Vout) × Iout. Vin × Iin isn't correct.
Option B: Incorrect. Input voltage divided by output current isn't correct - power dissipated = (Vin - Vout) × Iout. Vin / Iout isn't correct.
Option D: Incorrect. Output voltage multiplied by output current isn't correct - power dissipated = (Vin - Vout) × Iout. Vout × Iout isn't correct.
题目解析
The correct answer is C: Voltage difference from input to output multiplied by output current. Which of the following calculates power dissipated by a series linear voltage regulator is voltage difference from input to output multiplied by output current. Power = (Vin - Vout) × Iout. For amateur radio operators, this is important for power supply design. Understanding this helps when calculating power dissipation.
考试技巧
Power dissipated by series linear regulator = (Vin - Vout) × Iout. Think 'P'ower 'D'issipated = 'D'ifference 'V'oltage × 'O'utput 'C'urrent. Power = (Vin - Vout) × Iout. Not Vin × Iin, not Vin / Iout, not Vout × Iout - just (Vin - Vout) × Iout.
记忆口诀
Power dissipated by series linear regulator = (Vin - Vout) × Iout. Think 'P'ower = 'D'ifference × 'C'urrent. Power = (Vin - Vout) × Iout. Important for power supply design.
实际应用示例
Power dissipated by a series linear voltage regulator: It's calculated as voltage difference from input to output multiplied by output current. For example, if Vin=12V, Vout=5V, Iout=1A, then power = (12-5) × 1 = 7W. This is the calculation - (Vin - Vout) × Iout.
错误选项分析
Option A: Incorrect. Input voltage multiplied by input current isn't correct - power dissipated = (Vin - Vout) × Iout. Vin × Iin isn't correct. Option B: Incorrect. Input voltage divided by output current isn't correct - power dissipated = (Vin - Vout) × Iout. Vin / Iout isn't correct. Option D: Incorrect. Output voltage multiplied by output current isn't correct - power dissipated = (Vin - Vout) × Iout. Vout × Iout isn't correct.
知识点
Calculates power dissipated, Series linear voltage regulator, Voltage difference from input to output, Multiplied by output current
Verified Content
Question from official FCC Extra Class question pool. Explanation reviewed by licensed amateur radio operators.