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

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

Note: Formula to calculate 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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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.

Hint: The question asks about "individual currents" and while two answers mention current, only one answer--the correct one--uses the plural "currents".

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To solve for power use Ohm's Law \(I = E / R\) and Watt's Law \(P = I * E\), where \(I\) = current in amperes, \(E\) = voltage, \(R\) = resistance in ohms, and \(P\) = power in watts.

We know \(E = 400V\), and \(R = 800Ω\).

\[\text{Method 1:}\]

First solve for \(I\), using \(I = E/R\): \[400/800 = .5A\] Then solve for \(P\), using \(P = I*E\): \[.5 * 400 = 200\text{ watts.}\]

\[\text{Method 2:}\]

Combining the two equations gives us \(P = (E/R)* E\), or \(P = E^2/R\).

Solving for the given values:

\[400^2 / 800 = 200\text{ watts.}\]

Memory Tip: 200 doubled is 400. 400 doubled is 800.

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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

Memory Tip: 12x2=24. Move the decimal one place, 2.4.

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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

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

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

Memory tip: 7x1250 = a big number. 61 is bigger than 11. The bigger answer (61) that includes milli is correct. Milliamperes is in the question.

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The output Peak Envelope Power, or PEP, is 100 watts.

The question gives a peak-to-peak voltage of 200 V across a 50 ohm load. Since this is an AC sine wave, we first convert peak-to-peak voltage to peak voltage:

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

For a sine wave, RMS voltage is the peak voltage divided by \(\sqrt{2}\):

\[ \begin{align} V_{RMS} &= \frac{V_{peak}}{\sqrt{2}}\\ V_{RMS} &= \frac{100 V}{\sqrt{2}}\\ V_{RMS} &\approx 70.7 V \end{align} \]

Now use the power formula:

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

Substitute the RMS voltage and resistance:

\[ \begin{align} P &= \frac{70.7^2}{50}\\ P &\approx \frac{5000}{50}\\ P &\approx 100 W \end{align} \]

So the peak envelope power is approximately 100 watts.

Memory aids:

• Peak-to-peak voltage is twice the peak voltage.
• For a sine wave, \(V_{RMS} = \frac{V_{peak}}{\sqrt{2}}\).
• Power in a resistance can be found with \(P = \frac{E^2}{R}\).

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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

Silly: AC/DC Rocks My Socks (RMS)

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

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 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: \(120 \times 2.828 = 339.4\)

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Another way to look at it:

You may know \(\frac{1}{\sqrt{2}} = 0.707\) which is the factor used to multiply times peak voltage to obtain RMS voltage. To reverse the process, divide the RMS voltage by the already memorized 0.707 to obtain the peak voltage and double it to obtain peak-to-peak. \((\frac{120V}{0.707} \times 2) = 339.4\)

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Another way to look at it: \[V_\text{rms} = V_\text{peak} \times (.707)\] therefore: \[V_\text{peak} = \frac{V_\text{rms}}{.707} = \frac{120}{.707} = 169.73\] therefore: \[V_p-p = V_\text{peak} \times 2 = 169.73 \times 2 = 339.46\]

Memory Tip: The peak is the highest number. 339.4

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12 volts is the RMS voltage of a sine wave with a peak voltage of 17 volts.

For a sine wave, RMS voltage is found by multiplying the peak voltage by 0.707. This is the same as dividing by \(\sqrt{2}\):

\[ 0.707 \approx \frac{1}{\sqrt{2}} \]

So:

\[ \begin{align} V_{RMS} &= V_{peak} \times 0.707\\ V_{RMS} &= 17 V \times 0.707\\ V_{RMS} &\approx 12 V\\ \end{align} \]

So a sine wave with a peak voltage of 17 volts has an RMS voltage of about 12 volts.

Memory aids:

• For a sine wave, RMS voltage is about 70% of peak voltage.
• 12 and 17 both begin with 1.

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A transmission line loss of 1 dB means the output power is about 79.4% of the input power. That means the power loss is about 20.6%.

The power ratio formula for decibels is:

\[dB = 10 \times \log_{10}\left(\frac{P_f}{P_i}\right)\]

For a 1 dB loss, the change is negative:

\[\begin{align} -1 &= 10 \times \log_{10}\left(\frac{P_f}{P_i}\right)\\ -0.1 &= \log_{10}\left(\frac{P_f}{P_i}\right)\\ 10^{-0.1} &= \frac{P_f}{P_i}\\ \frac{P_f}{P_i} &\approx 0.794\\ \end{align} \]

So after a 1 dB loss, about 79.4% of the power remains.

The percentage of power lost is:

\[\begin{align} 100\% - 79.4\% &= 20.6\%\\ \end{align}\]

So the power loss is about 20.6%.

Memory aids:

• A 1 dB loss leaves about 79% of the power.
• A 1 dB loss is about a 21% power loss.
• A 3 dB loss is about half power.
• Three 1 dB losses should be close to one 3 dB loss:

\[ \begin{align} 0.794^3 &\approx 0.5\\ \end{align} \]

Note: 25.9% is the percentage increase for a 1 dB gain, not the percentage loss for a 1 dB loss.

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The key word here is "unmodulated."

A modulation envelope shows how a carrier wave's amplitude changes over time, the envelope power is the power value of that envelope, and the peak envelope power is the highest power value within some timeframe.

However, if a carrier is unmodulated, its power isn't changing, which means its modulation envelope is flat. And as with any unchanging value, its peak equals its average, and the ratio of two equal values is \[1:1 = 1/1 = 1.00\]

(Note that as power varies directly with amplitude\(^2\), it's convenient to think of the envelope as representing both.)

(Note also that the question asks about the ratio of two power values, not the values themselves, so RMS calculations aren't involved here.)

Hint: There’s only one (1.00) answer.

Useful further reading:

• https://www.rfglobalnet.com/doc/basics-of-power-measurement-average-or-peak-0001
• https://resources.pcb.cadence.com/blog/2020-peak-envelope-power-and-your-rf-signal-chain-simulations

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

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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}

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

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Silly Hint: The correct answer is the only one that has both a 2 (as in 1200 watts) and a 5 (as in 50 ohms).
Or Think about a bathroom heater "240 Volts"

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1060 watts is the output PEP (Peak Envelope Power) 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.

Silly hint: 1060 is both in the question and answer

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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}

Silly hint: 500 has 3 digits whole digits. 625 is the only whole number with 3 digits

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