Free Ham Radio Extra Class Questions and Answers

Question 1

Which ionospheric layer is the nearest to Earth's surface?

Accepted Answer

The D layer

Answer

The D layer is the lowest region of the Earth's ionosphere, typically found at altitudes between 50 and 90 kilometers during daylight hours. It is primarily responsible for absorbing lower frequency radio waves, especially during the day, which can limit long-distance communication on those bands. The E and F layers are located at progressively higher altitudes above the D layer.

Question 2

What benefit does employing a ferrite core toroidal inductor provide?

Accepted Answer

All of the above

Answer

Ferrite core toroidal inductors offer several significant advantages in electronic circuits. Their toroidal shape effectively contains most of the magnetic field within the core, minimizing external interference and improving efficiency. The high permeability of ferrite material allows for the creation of large inductance values in a compact size, and different ferrite compositions can be chosen to optimize the core's magnetic properties for specific frequency ranges, making them highly versatile.

Question 3

In a 146.52 MHz FM phone transmitter with a 5 kHz dispersion, what is the frequency deviation for a 12.21 MHz reactance modulated oscillator?

Accepted Answer

416.7 Hz

Answer

In an FM transmitter, the frequency deviation produced by the reactance modulated oscillator is multiplied to reach the final output frequency and deviation. First, calculate the frequency multiplication factor by dividing the final output frequency (146.52 MHz) by the oscillator frequency (12.21 MHz), which is 12. Then, divide the desired final frequency dispersion (5 kHz) by this multiplication factor (12) to find the oscillator's deviation, which is approximately 416.7 Hz.

Question 4

If an oscilloscope records 200 volts peak-to-peak across a 50 ohm dummy load attached to the transmitter output, what is the output PEP from the transmitter?

Accepted Answer

100 watts

Answer

To calculate the Peak Envelope Power (PEP) from a peak-to-peak voltage measurement across a resistive load, first determine the peak voltage. A 200-volt peak-to-peak signal means the peak voltage is 100 volts. Then, use the formula P = Vp^2 / (2 * R), where Vp is the peak voltage and R is the resistance. Plugging in the values, (100V)^2 / (2 * 50 ohms) equals 10000 / 100, resulting in an output PEP of 100 watts.

Question 5

What is a 22,000 picofarad (pF) capacitor's value in nanofarads (nF)?

Accepted Answer

22

Answer

To convert picofarads (pF) to nanofarads (nF), you need to remember that 1 nanofarad is equal to 1000 picofarads. Therefore, to convert 22,000 picofarads to nanofarads, you divide 22,000 by 1000. This calculation yields a value of 22 nanofarads.

HAM - Radio Extra Class Test Practice Test

HAM - Radio Extra Class Test Practice Test

Free Ham Radio Extra Class Questions and Answers