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Subelement G5
Electrical Principles
Section G5B
The decibel; current and voltage dividers; electrical power calculations; sine wave root-mean-square (RMS) values; PEP calculations
What dB change represents a factor of two increase or decrease in power?
  • Approximately 2 dB
  • Correct Answer
    Approximately 3 dB
  • Approximately 6 dB
  • Approximately 12 dB

(B). A two-times increase or decrease in power results in a change of approximately 3 decibels.


The logarithmic decibel scale is used to measure changes in power or signal strength.

Note: Formula for change in Power in Decibels

\[\Delta\text{ dB} = 10 \times \log_{10} \bigg ( \frac{P_2}{P_1} \bigg )\]

Where:
\(P_1 =\) reference power
\(P_2 =\) power being compared.

If we plug in the values given in this question we get:

\begin{align} \Delta\text{ dB} &= 10 \times \log_{10} \bigg ( \frac{P_2}{P_1} \bigg )\\ &= 10 \times \log_{10} \bigg ( \frac{2}{1} \bigg )\\ &= 10 \times \log_{10} (2)\\ &= 10 \times 0.3\\ &= 3\text{ dB}\\ \end{align}

Hint: Watch out for (C). Amateur radio uses an S scale to measure signal strength, and a change of 1 S unit corresponds to a four-times increase in power or a 6 dB change, but this is NOT what they are asking for in this question. Don't get fooled.

For more info see Wikipedia: Decibel (dB)

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Tags: arrl chapter 4 arrl module 10

How does the total current relate to the individual currents in each branch of a purely resistive parallel circuit?
  • It equals the average of each branch current
  • It decreases as more parallel branches are added to the circuit
  • Correct Answer
    It equals the sum of the currents through each branch
  • It is the sum of the reciprocal of each individual voltage drop

Current doesn't get lost. It gets split up across all of the branches of the circuit, but all the currents in all the branches add up to the total current.

The equation for the total current of a parallel circuit is:

\[\text{I}_{total} = \text{I}_1 + \text{I}_2 + \ldots + \text{I}_n\]

For an excellent, intuitive explanation of this, see the video from the Khan Academy on Kirchhoff's current law.

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Tags: arrl chapter 4 arrl module 12

How many watts of electrical power are used if 400 VDC is supplied to an 800 ohm load?
  • 0.5 watts
  • Correct Answer
    200 watts
  • 400 watts
  • 3200 watts

(B). The amount of electrical power used if 400 VDC (volts direct current) is supplied to an 800-ohm load is 200 watts. Use "Ohm's law circle" and the Joule's Law "Power Circle" equations to solve the problem: I = Current in Amperes, E = Voltage, R = Resistance in Ohms, P = Power in Watts.

We know E = 400 Volts, and R = 800 ohms

To solve for power use the Ohm's Law Equation I = E / R and combine it with the power equation: P = I x E to give us:

P = E x (E / R) which is the same as P = (E x E) / R or P = E2 / R.

So solving this question: P = 400 x (400 / 800) = 200 Watts or P = 400^2 / 800 = 200 Watts

=========== NOTE: Error correction follows ===========

The formula required P = E x (E / R) which is the same as P = (E x E) / R or P = E squared / R. is the correct formula, however, the math illustrated does provide the answer of 200 watts which is correct. Just the math portion of the illustration above will not factor a result of 200 watts.

POWER = VOLTS * VOLTS / OHMS P= 400 * 400 /800 is what needs to be solved.

THE SOLUTION:

  1. 400 * 400=160,000.

  2. 160,000 / 800 = 200.

Therefore the answer is 200 watts of power is used.

Edited by. greeneggsand

Simple: Draw the E/IR chart and the P/IE chart then you can do this in your head.

(E)400volts/(R)800ohms =(I) .5amps

IxE = P I(.5) x E(400) = 200w

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Tags: arrl chapter 4 arrl module 10

How many watts of electrical power are used by a 12 VDC light bulb that draws 0.2 amperes?
  • Correct Answer
    2.4 watts
  • 24 watts
  • 6 watts
  • 60 watts

(A). The amount of power (watts) used by a 12-VDC (volts direct current) light bulb that draws 0.2 amperes is 2.4 watts. The power circle equation based on Ohm's law shows that P = I x E, where P is power in Watts, I is current in Amperes, and E is energy in Volts.

So for this question: P = I x E

P = (0.2 amperes) x (12 volts) = 2.4 watts

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Tags: arrl chapter 4 arrl module 10

How many watts are dissipated when a current of 7.0 milliamperes flows through a 1250 ohm resistance?
  • Correct Answer
    Approximately 61 milliwatts
  • Approximately 61 watts
  • Approximately 11 milliwatts
  • Approximately 11 watts

Approximately 61 milliwatts are dissipated when a current of 7.0 milliamperes flows through 1.25 kilohms.

By combining the Ohm's Law equations and the power circle equations we can solve the problem.

  • \(P\) = Power in Watts
  • \(I\) = Current in Amperes
  • \(E\) = Energy in Volts
  • \(R\) = Resistance in Ohms

We are given:

\(I_\text{current}\) \(=\) \(7.0 \text{milliamperes}\) \(=\) \(0.007 \text{Amperes}\)

\(R_\text{resistance}\) \(=\) \(1.25 \text{kilohms}\) \(=\) \(1250 \text{Ohms}\)

To solve for \(P\), use the power equation \(P = I \times E\) and combine it with the equation from Ohm's Law \(E = I \times R\).

This gives us: \(P = I \times (I \times R)\) \(=\) \(I^2 \times R\)

So for this question:

\(P = 0.007 A \times (0.007 A \times 1250 Ω)\) \(=\) \(0.061 \text{Watts}\) \(=\) \(61 \text{milliwatts}\), or

\(P = (0.007 A)^2 \times 1250 Ω\)=\(0.061 \text{Watts}\)=\(61 \text{milliwatts}\)


Alternate hint: \(\bigg(P = \frac{I^2 \times R}{1,000}\bigg)\) (divide by 1,000 because it’s measured in milliamperes)

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Tags: arrl chapter 4 arrl module 10

What is the output PEP from a transmitter if an oscilloscope measures 200 volts peak-to-peak across a 50 ohm dummy load connected to the transmitter output?
  • 1.4 watts
  • Correct Answer
    100 watts
  • 353.5 watts
  • 400 watts

The output Peak Envelope Power (PEP) is 100 watts.


This problem can be solved using Joule's Law (also known as the power law):

\[P = E \times I\]

There are two problems here, though.

First, the question provides values for voltage and resistance, but not current. That's OK, though, since the current can be derived using Ohm's Law.

Second, the question provides a value for peak-to-peak voltage (\(V_{PP}\)) and assumes an alternating current (AC) in the form of sine wave. \(V_{PP}\) can't be plugged straight into the power law as is. It will first have to be converted to peak voltage (\(V_{peak}\)) and then to the root mean square voltage (\(V_{RMS}\)) value.


To get a version of the power law that will solve this problem, take Ohm's law, rearranged to define current:

\[I = \frac{E}{R}\]

and plug this into the power law in place of I:

\[P = E \times \frac{E}{R}\]

\(E \times E\) can be written as \(E^2\):

\[P = \frac{E^2}{R}\]

This is the power equation used for direct current (DC). But that brings us to the second problem. Because we're working with AC, in order to plug a value of \(E\) into this equation, the voltage needs to be expressed in terms of \(V_{RMS}\).

The peak voltage is simply the peak-to-peak voltage divided by two:

\begin{align} V_{peak}& = \frac{V_{PP}}{2}\\ &=\frac{200\text{ V}}{2}\\ &=100\text{ V}\\ \end{align}

Next, the root mean square voltage for a sine wave can be calculated from the peak voltage as follows:

\begin{align} V_{RMS} & = \frac{V_{peak}}{\sqrt{2}}\\ & = \frac{100\text{ V}}{\sqrt{2}}\\ \end{align}

Now we can calculate the PEP using the power law, as derived above:

\begin{align} P & = \frac{E^2}{R}\\ &= \frac{\Big (\frac{100}{\sqrt{2}} \Big )^2}{50\ \Omega}\\ &= \frac{\frac{10000}{2}}{50}\\ &= \frac{5000}{50}\\ &= 100\text{ W}\\ \end{align}

Hint: If you calculate \(V_{RMS}\) directly for 100 volts you will get a value for E of approximately 70.7 volts. Squaring that and then dividing it by R (in our example, 50 ohms) will give you 99.9698 watts, which is approximately 100 watts.

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Tags: arrl chapter 4 arrl module 11

What value of an AC signal produces the same power dissipation in a resistor as a DC voltage of the same value?
  • The peak-to-peak value
  • The peak value
  • Correct Answer
    The RMS value
  • The reciprocal of the RMS value

(C). The RMS (root mean square) value of an AC signal results in the same power dissipation as a DC voltage of the same value. The RMS value is useful as an average value for the voltage in an AC circuit throughout the alternating wave cycle, as if the current were a constant as in DC.

For more info see Wikipedia: RMS, or root mean square

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Tags: arrl chapter 4 arrl module 11

What is the peak-to-peak voltage of a sine wave with an RMS voltage of 120.0 volts?
  • 84.8 volts
  • 169.7 volts
  • 240.0 volts
  • Correct Answer
    339.4 volts

For a sine wave the RMS voltage of a sine wave is approximately \(0.7 \times \text{the peak voltage}\).

Looking at the possible answers, which are all peak-to-peak: Divide them in half to know the peak voltage, and only one answer is still high enough to possibly be correct: \(\frac{339.4}{2} = 169.7\). Multiply that by \(0.7\) to get approximately \(120V\).

Alternatively, for a sine wave the peak-to-peak voltage = \(2\sqrt{2} \times \text{RMS voltage} = 2.828 \times \text{RMS voltage}\). So: \(120 \times 2.828 = 339.4\)


Alt Hint: (\(V \times 0.707\)) \(=\) RMS Voltage (sine), but since it asks for peak-to-peak, multiply result by \(4\). \((120V \times 0.707) \times 4 = 339.4\)

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Tags: arrl chapter 4 arrl module 11

What is the RMS voltage of a sine wave with a value of 17 volts peak?
  • 8.5 volts
  • Correct Answer
    12 volts
  • 24 volts
  • 34 volts

12 volts is the RMS (i.e. Root Mean Square) voltage of a sine wave with a value of 17 volts peak.

To find the RMS voltage, multiply the peak voltage by 0.707 (which is 1/sqrt(2), so it's the same as dividing by the square root of 2).

RMS voltage = peak voltage x 0.707

So for this problem RMS voltage = 17 volts x 0.707 = 12 volts.

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Tags: arrl chapter 4 arrl module 11

What percentage of power loss would result from a transmission line loss of 1 dB?
  • 10.9 percent
  • 12.2 percent
  • Correct Answer
    20.6 percent
  • 25.9 percent

(C). A percentage power loss of 20.5% would result from a transmission line power loss of 1 dB.

Given
\begin{align} P_i &= Initial\ Power\\ P_f &= Final\ Power \end{align}

The generalized formula to compute the power change (i.e. loss or gain) in decibels (dB) is:
\[\Delta\ \text{dB} = 10 \times \log_{10} \bigg (\frac{P_f}{P_i} \bigg )\]

In this problem, the power change is given as "a transmission line loss of 1 dB", which we can plug into our equation:
\[-1\ \text{dB} = 10 \times \log_{10} \bigg (\frac{P_f}{P_i} \bigg )\]

Let's isolate the variables \(P_i\) and \(P_f\). First, divide both sides by 10:
\[\frac{-1}{10}\ \text{dB} = \log_{10} \bigg (\frac{P_f}{P_i} \bigg )\]

Exponent (i.e. raise) 10 to the power of both sides of the equation:
\[10^\frac{-1}{10} = 10^{\log_{10} \Big (\frac{P_f}{P_i} \Big )}\]

Simplify:
\[10^\frac{-1}{10} = \bigg (\frac{P_f}{P_i} \bigg )\]

The problem asks us to calculate the "percentage of power loss", which will be the absolute value of % power change.
\begin{align} \%_{power\ change} &= 100 \times \frac{P_f - P_i}{P_i}\\ &= 100 \times \bigg (\frac{P_f}{P_i} - 1 \bigg ) \end{align}

Plug in the calculated value \(\Big (\frac{P_f}{P_i} = 10^\frac{-1}{10} \Big )\) :
\begin{align} \%_{power\ change} &= 100 \times (10^\frac{-1}{10} - 1 )\\ &= -20.6\ \% \end{align}

So:
\begin{align} \%_{power\ loss} &= | \%_{power\ change} |\\ &= |-20.6\ \%|\\ &= 20.6\ \%\\ &\approx 20.5\ \% \end{align}

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Tags: arrl chapter 4 arrl module 10

What is the ratio of peak envelope power to average power for an unmodulated carrier?
  • 0.707
  • Correct Answer
    1.00
  • 1.414
  • 2.00

(B). The ratio of peak envelope power to average power for an unmodulated carrier is 1.00. An unmodulated carrier gives a straight RF carrier wave, so the average and peak envelope voltages are the same, so the ratio is 1:1 or 1.00.


Hint: PEP is same as Average Power (unmodulated)

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Tags: arrl chapter 4 arrl module 11

What would be the RMS voltage across a 50 ohm dummy load dissipating 1200 watts?
  • 173 volts
  • Correct Answer
    245 volts
  • 346 volts
  • 692 volts

(B). The RMS voltage across a 50-ohm dummy load dissipating 1200 watts is 245 volts.


You are given:
\begin{align} Power (P) &= 1200\text{ watts}\\ Resistance (R) &= 50\ \Omega \end{align}

Combining the power circle equation \(P = I \times E\) and the Ohm's Law equation \(I = \frac{E}{R}\) gives the equation: \[P = \frac{ E^2 }{ R\ \ }\]

Rearranging to solve for \(E\):

\[E = \sqrt{ P \times R }\]

Using the values in this question:

\begin{align} E &= \sqrt{ 1200\text{ W} \times 50\ \Omega }\\ &= \sqrt{ 60000 }\text{ V}\\ &= 100\sqrt{ 6 }\text{ V}\\ &= 245\text{ V} \end{align}


Alternate Hint: \(E= \sqrt{P \times R}\). \(1200 \times 50 = 60,000\); (\(\sqrt{60,000} = 245\))

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Tags: arrl chapter 4 arrl module 11

What is the output PEP of an unmodulated carrier if an average reading wattmeter connected to the transmitter output indicates 1060 watts?
  • 530 watts
  • Correct Answer
    1060 watts
  • 1500 watts
  • 2120 watts

(B). 1060 watts is the output PEP of an unmodulated carrier if an average reading wattmeter connected to the transmitter output indicates 1060. The average power and the PEP are the same value for an unmodulated carrier, so the values are the same.

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Tags: arrl chapter 4 arrl module 11

What is the output PEP from a transmitter if an oscilloscope measures 500 volts peak-to-peak across a 50 ohm resistive load connected to the transmitter output?
  • 8.75 watts
  • Correct Answer
    625 watts
  • 2500 watts
  • 5000 watts

(B). 625 watts is the output PEP from a transmitter if an oscilloscope measures 500 volts peak-to-peak across a 50-ohm resistor connected to the transmitter output.

Use the combined power and Ohm's law equation:

\[P = \frac{E^2}{R}\]

where: \begin{align} P &= \text{Peak Envelope Power}\\ E &= \text{Root Mean Square (RMS) Voltage} \end{align}

First find the \(E_{RMS}\):

\[E_{RMS} = \frac{\text{Peak Envelope Voltage}}{\sqrt{2}}\]

To find the Peak Envelope Voltage (PEV): \begin{align} \text{PEV} &= \frac{\text{Peak-to-Peak Voltage}}{2}\\ &= \frac{500\ \text{V}}{2}\\ &= 250\ \text{V} \end{align}

So now the Peak Envelope Power (PEP) output in this case can be calculated as:

\begin{align} P_{PEP} &= \frac{(E_{RMS})^2}{R}\\ &= \frac{ \Big ( \frac{ 250\ \text{V}}{ \sqrt{2} } \Big ) ^2}{50\ \Omega}\\ &= 625\ \text{W} \end{align}

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