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Subelement T5
Electrical principles, math for electronics, electronic principles, Ohm's Law
Section T5B
Math for electronics; decibels, electrical units and the metric system
How many milliamperes is 1.5 amperes?
  • 15 milliamperes
  • 150 milliamperes
  • Correct Answer
    1,500 milliamperes
  • 15,000 milliamperes

The prefix "milli" means "one thousandth", so move the decimal to the right 3 places (or multiply by 1,000$) to go from amperes to milliamperes.

\[1.5 \times 1,000 = 1,500\]

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 maxhamstudy. Register to edit

Tags: electrical current math

What is another way to specify a radio signal frequency of 1,500,000 hertz?
  • Correct Answer
    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}

Last edited by meisinger. Register to edit

Tags: frequencies math

How many volts are equal to one kilovolt?
  • One one-thousandth of a volt
  • One hundred volts
  • Correct Answer
    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.

Last edited by kd7bbc. Register to edit

Tags: electromotive force (voltage) math

How many volts are equal to one microvolt?
  • Correct Answer
    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.

One one-thousandth of a volt is denoted with the millivolt. Also if you think of it this way milli is like saying million, so 1 out of 1 million

Last edited by trey.lovell. Register to edit

Tags: electromotive force (voltage) math

Which of the following is equivalent to 500 milliwatts?
  • 0.02 watts
  • Correct Answer
    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 \(\frac{500}{1000} = 0.5 \text{ watts}\)

Last edited by kd7bbc. Register to edit

Tags: electrical power math

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
  • Correct Answer
    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.

Last edited by kd7bbc. Register to edit

Tags: electrical current math

If a frequency readout 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
  • Correct Answer
    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.

Last edited by rjstone. Register to edit

Tags: frequencies math

How many microfarads are 1,000,000 picofarads?
  • 0.001 microfarads
  • Correct Answer
    1 microfarad
  • 1000 microfarads
  • 1,000,000,000 microfarads

Remember the order of your metric units!

  • 1 farad = 1,000 millifarads (mF)
  • 1 millifarad = 1,000 microfarads (μF)
  • 1 microfarad = 1,000 nanofarads (nF)
  • 1 nanofarad = 1,000 picofarads (pF)

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: \(10^6 pF\) \(=\) \(10^6 \times 10^-12 F\) = \(10^{6-12} F\) \(=\) \(10^{-6} F = 1 μF\)

You only have to remember that every metric prefix has three 0´s. Write three 0´s under every prefix. Make a sheet like this:

x| milli | micro | nano | pico |
x | x x x | x x 1 | 000 | 000 |

You can read it from the sheet: 1 microfarad.

Last edited by markus13el1. Register to edit

Tags: inductance math

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

\[L\text{dB} = 10\times \log_{10} \left(\frac{P_1}{P_0}\right)\]

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) \\=\log_{10}(2^{10}) \\=\log_{10}(1024)\\ \approx3\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 braian. Register to edit

Tags: transmit power math

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
  • Correct Answer
    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.

Last edited by kd7bbc. Register to edit

Tags: transmit power

What is the approximate amount of change, measured in decibels (dB), of a power increase from 20 watts to 200 watts?
  • Correct Answer
    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 \(10\) =]

Last edited by masteraviator. Register to edit

Tags: transmit power math

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