Ham Radio General Class Test 2026 – 400 Free Practice Questions to Pass the Exam

Session length

1 / 785

How is the efficiency of an RF power amplifier determined?

Divide the DC input power by the DC output power

Divide the RF output power by the DC input power

The efficiency of an RF power amplifier is determined by the ratio of the RF output power to the DC input power. This ratio indicates how effectively the amplifier converts the DC power it receives into RF output power.

In practical terms, an amplifier takes a certain amount of DC (direct current) power and converts it to RF (radio frequency) signals. The efficiency is expressed as a percentage, calculated by taking the RF output power and dividing it by the DC input power, then multiplying by 100. A higher efficiency means that more of the input power is being converted into useful RF output rather than being lost as heat or other forms of energy.

Understanding this relationship is crucial for evaluating amplifier performance, as it helps radio operators and engineers determine how effectively their equipment operates in terms of power usage and thermal management.

Get further explanation with Examzify DeepDiveBeta

Multiply the RF input power by the reciprocal of the RF output power

Add the RF input power to the DC output power

Next Question
Subscribe

Get the latest from Examzify

You can unsubscribe at any time. Read our privacy policy