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Subelement T5
Electrical principles: math for electronics; electronic principles; Ohm's Law
Section T5B
Math for electronics: conversion of electrical units; decibels; the metric system
How many milliamperes is 1.5 amperes?
• 15 milliamperes
• 150 milliamperes
• 1500 milliamperes
• 15,000 milliamperes

The prefix "milli" means "one thousandth", so move the decimal to the right 3 places (or multiply by $1000$) to go from amperes to milliamperes.

$1.5 \times 1000 = 1,500$

A quick way to check each answer is to remember that to convert milliamps to amps, just change the comma to a decimal point. $1,500mA = 1.500A$

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Tags: electrical current math arrl chapter 2 arrl module 2

What is another way to specify a radio signal frequency of 1,500,000 hertz?
• 1500 kHz
• 1500 MHz
• 15 GHz
• 150 kHz

Pay attention when you get these questions; approximately 30% of applicants who saw this question over a 2 year period answered 1500 MHz instead of 1500 kHz, and I'm pretty sure that they actually did know how to do the conversion. It would of course be 1.5 MHz, but it is certainly not 1500 MHz.

\begin{align} 1,000\text{ Hz} = 1\text{ kHz} \end{align} \begin{align} \frac{1,500,000\text{ Hz}}{1\text{ kHz }} = 1500\text{ kHz} \end{align}

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Tags: frequencies math arrl chapter 2 arrl module 2

How many volts are equal to one kilovolt?
• One one-thousandth of a volt
• One hundred volts
• One thousand volts
• One million volts

The prefix "kilo", commonly used in all metric forms of measurement, means "thousand". Thus, a "kilovolt" is a thousand volts.

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Tags: electromotive force (voltage) math arrl chapter 2 arrl module 2

How many volts are equal to one microvolt?
• One one-millionth of a volt
• One million volts
• One thousand kilovolts
• One one-thousandth of a volt

"micro" is a prefix in the metric system meaning "one-millionth". Thus, a microvolt is one millionth ($\frac{1}{1,000,000}$) of a volt.

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Tags: electromotive force (voltage) math arrl chapter 2 arrl module 2

Which of the following is equal to 500 milliwatts?
• 0.02 watts
• 0.5 watts
• 5 watts
• 50 watts

$1000$ milliwatts = $1$ watt (remember that there are $10$ mm in $1$ cm and $100$ cm in a meter; or $1000$ millimeters in a meter)

-or-

$1000 \text{ milliwatts} = 1$ watt, so $\frac{500}{1000} = 0.5 \text{ watts}$

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Tags: electrical power math arrl chapter 2 arrl module 2

If an ammeter calibrated in amperes is used to measure a 3000-milliampere current, what reading would it show?
• 0.003 amperes
• 0.3 amperes
• 3 amperes
• 3,000,000 amperes

milliamp (milliampere) = $\frac{1}{1000}$ of amp(ampere)

Remember your standard metrics; the ammeter is reading amperes, so $1000$ milliamperes is $1$ ampere, and $3000$ milliamperes is $3$ amperes.

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Tags: electrical current math arrl chapter 2 arrl module 2

If a frequency display calibrated in megahertz shows a reading of 3.525 MHz, what would it show if it were calibrated in kilohertz?
• 0.003525 kHz
• 35.25 kHz
• 3525 kHz
• 3,525,000 kHz

MHz is 1,000 times more than kHz or...

$1\text{ MHz} = 1,000\text{ kHz}$

M is Mega = $10^6$

k is Kilo = $10^3$

Therefore: $\frac{3.525\text{ MHz }\times 1000\text{ kHz}}{1\text{ MHz}} = 3525\text{ KHz}$

Since we are multiplying by 1 we do not change the value of what is represented. $\frac{1000\text{ kHz}}{1\text{ MHz}} = 1$.

This can be applied for any unit conversions.

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Tags: frequencies math arrl chapter 2 arrl module 2

Remember the order of your metric units!

Therefore $1,000,000$ picofarads = $1,000$ nanofarads = 1 microfarad

Or group using decimals: $1,000,000 pF \times .000 000 000 001 F / pF$ pF cancels out and you get $.000 001 F = 1 μF$

And if you're feeling scientific use exponents: $10^6 pF$ $=$ $10^6 \times 10^-12 F$ = $10^{6-12} F$ $=$ $10^{-6} F = 1 μF$

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Tags: inductance math arrl chapter 2 arrl module 2

What is the approximate amount of change, measured in decibels (dB), of a power increase from 5 watts to 10 watts?
• 2 dB
• 3 dB
• 5 dB
• 10 dB

When dealing with decibels, every $3\text{dB}$ of gain doubles the power, and every $3\text{dB}$ of loss halves it. So, $5$ watts to $10$ watts is twice the power, so it is $3\text{dB}$. $5$ watts to $20$ watts would be four times the power, or $2 \times 2 \times P$ (where $P$ is power), or $6\text{dB}$. $9\text{dB}$ would be $40$ watts, $12\text{dB}$ would be $80$ watts, etc.

Similarly, $-3\text{dB}$ would be $2.5$ watts, since $-3\text{dB}$ is half the power, and $-6\text{dB}$ would be half of that, $-9\text{dB}$ half of that, etc.

If you're interested, the formula to calculate the gain (or loss) of power in decibels is:

$L\text{dB} = 10\times \log_{10} (\frac{P_1}{P_0})$

Where $L$ is gain in $\text{dB}$, $P_1$ is the new value and $P_0$ is the old. In this case,

\begin{align} 10 \times \log_{10} \frac{10}{5} = 10 \times \log_{10}(2) \\= 10 \times 0.3 \\= 3\text{dB} \end{align}

This is difficult to calculate without a good calculator, though, and the $3\text{dB}$ = twice rule is easier to remember.

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Tags: transmit power math arrl chapter 4 arrl module 8

What is the approximate amount of change, measured in decibels (dB), of a power decrease from 12 watts to 3 watts?
• -1 dB
• -3 dB
• -6 dB
• -9 dB

When dealing with decibels, every $3 dB$ of gain doubles the power, and every $3 dB$ of loss halves it. So, $5$ watts to $10$ watts is twice the power, so it is $3 dB$. $5$ watts to $20$ watts would be four times the power, or $2 \times 2 \times P$ (where $P$ is power), or $6dB$. $9dB$ would be $40$ watts, $12dB$ would be $80$ watts, etc.

Similarly, $-3dB$ would be $2.5$ watts, since $-3dB$ is half the power, and $-6dB$ would be half of that, $-9dB$ half of that, etc.

In this case, $12$ watts to $3$ watts is $\frac{1}{4}$ of the power, or $\frac{1}{2} \times \frac{1}{2}$ of the power, which is $-6dB$. Since they're only asking the amount of change, and not the direction, the answer is just $6 dB$ (of loss).

If you're interested, the formula to calculate the gain (or loss) of power in decibels is:

$Ldb = 10 \times \log (\frac{P_1}{P_0})$

Where $\log$ is the log base 10 and $P_1$ is the new value and $P_0$ is the old. In this case, $10 \times \log (\frac{3}{12})$ $=$ $10 \times \log(\frac{1}{4})$ $=$ $10 \times -0.6$ $=$ $-6dB$. This is difficult to calculate without a good calculator, though, and the $3dB$ = twice rule is easier to remember.

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Tags: transmit power arrl chapter 4 arrl module 8

What is the amount of change, measured in decibels (dB), of a power increase from 20 watts to 200 watts?
• 10 dB
• 12 dB
• 18 dB
• 28 dB

The formula to calculate the gain (or loss) of power in decibels is:

$Ldb = 10 \times \log(\frac{P_1}{P_0})$

Where log is the log base 10 and $P_1$ is the new value and $P_0$ is the old. So in this case, $10 \times \log(\frac{200}{20})$ $=$ $10 \times \log(10)$ $=$ $10dB$. ($\log_{10} (10) = 1$).

If you have a hard time remembering that formula, though, there is an easier to remember rule: every $3dB$ of gain doubles the power, and every $-3dB$ of gain (or $3dB$ of loss) halves it. So, looking at this, $3dB$ of gain would be $20 \times 2 = 40$. $6 dB$ would be $20 \times 2 * 2 = 80$. $9dB$ would be $20 \times 2 \times 2 \times 2 = 160$, and $12dB$ would be $20 \times 2 \times 2 \times 2 \times 2 = 320$, which is too high, so you know it isn't $12$ but it's more than $9$; it must be $10$ =]

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Tags: transmit power math arrl chapter 4 arrl module 8

Which of the following frequencies is equal to 28,400 kHz?
• 28.400 MHz
• 2.800 MHz
• 284.00 MHz
• 28.400 kHz

To convert kHz to Mhz, move the decimal point three positions to the left. With 28,400 KHz the decimal point is assumed to be all the way to the right of 28400.

So to begin with, we have 28400. kHz (I took out the comma and put in the trailing decimal point to make it easier to see what's going on). So moving the decimal point three positions to the left, it ends up between the $28$ and the $400$, so $28.400$ MHz.

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Tags: arrl chapter 3 arrl module 2

If a frequency display shows a reading of 2425 MHz, what frequency is that in GHz?
• 0.002425 GHz
• 24.25 GHz
• 2.425 GHz
• 2425 GHz

To convert from MHz to GHz, move the decimal point three positions to the left.

Starting with $2425.0$ MHz, move the decimal point three positions to the left, ending up between the $2$ and $425$, so $2.425$ GHz.

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Tags: arrl chapter 2 arrl module 2