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. \(1,500mA = 1.500A\)
Last edited by kd7bbc. Register to edit
Tags: electrical current math arrl chapter 3 arrl module 5
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}
Last edited by meisinger. Register to edit
Tags: frequencies math arrl chapter 3 arrl module 5
The prefix "kilo", commonly used in all metric forms of measurement, means "thousand". Thus, a "kilovolt" is a thousand volts.
Last edited by kd7bbc. Register to edit
Tags: electromotive force (voltage) math arrl chapter 3 arrl module 5
"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.
Last edited by kd7bbc. Register to edit
Tags: electromotive force (voltage) math arrl chapter 3 arrl module 5
\(1000\) 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 \(\frac{500}{1000} = 0.5 \text{ watts}\)
Last edited by kd7bbc. Register to edit
Tags: electrical power math arrl chapter 3 arrl module 5
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.
Last edited by kd7bbc. Register to edit
Tags: electrical current math arrl chapter 3 arrl module 5
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.
Last edited by rjstone. Register to edit
Tags: frequencies math arrl chapter 3 arrl module 5
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: \(10^6 pF\) \(=\) \(10^6 \times 10^-12 F\) = \(10^{6-12} F\) \(=\) \(10^{-6} F = 1 μF\)
Last edited by kd7bbc. Register to edit
Tags: inductance math arrl chapter 3 arrl module 5
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.
Last edited by rjstone. Register to edit
Tags: transmit power math arrl chapter 4 arrl module 9
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.
Last edited by kd7bbc. Register to edit
Tags: transmit power arrl chapter 4
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 \(10\) =]
Last edited by masteraviator. Register to edit
Tags: transmit power math arrl chapter 4 arrl module 9
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.
Last edited by kd7bbc. Register to edit
Tags: arrl chapter 3 arrl module 5
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.
Last edited by kd7bbc. Register to edit
Tags: arrl chapter 3 arrl module 5