Deep Dive: E7D13
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.
Why Other Answers 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.
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 Example
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.
Source & Coverage
Question Pool: 2024-2028 Question Pool
Subelement: E7D
Reference: 2024-2028 Question Pool · E7 - Practical Circuits
Key Concepts
Verified Content
Question from the official FCC Extra Class pool. Explanation reviewed by licensed amateur radio operators and mapped to the E7D topic.