Correct answer: 20 dB

Voltage gain in decibels is calculated using:

\[ \mathrm{Gain(dB)} = 20 \log_{10}\!\left(\frac{V_{out}}{V_{in}}\right) \]

Here, the voltage increases from \(V_{in} = 1\ \mathrm{V_{rms}}\) to \(V_{out} = 10\ \mathrm{V_{rms}}\), giving a ratio of:

\[ \frac{V_{out}}{V_{in}} = 10 \]

Substituting:

\[ \mathrm{Gain} = 20 \log_{10}(10) = 20 \times 1 = 20\ \mathrm{dB} \]

• 3 dB corresponds to roughly a 1.41× voltage increase.
• 6 dB corresponds to roughly a 2× voltage increase.
• 10 dB corresponds to about a 3.16× voltage increase.

Therefore, increasing the voltage from 1 V rms to 10 V rms is a 20 dB increase.

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Correct answer: 40 dB

Voltage gain in decibels is calculated using:

\[ \mathrm{Gain(dB)} = 20 \log_{10}\!\left(\frac{V_{out}}{V_{in}}\right) \]

Here, the voltage increases from \(V_{in} = 1\ \mathrm{V_{rms}}\) to \(V_{out} = 100\ \mathrm{V_{rms}}\), giving a ratio of:

\[ \frac{V_{out}}{V_{in}} = 100 \]

Substituting:

\[ \mathrm{Gain} = 20 \log_{10}(100) = 20 \times 2 = 40\ \mathrm{dB} \]

• 10 dB corresponds to about a 3.16× voltage increase.
• 20 dB corresponds to a 10× voltage increase.
• 100 dB would require an extremely large voltage ratio far greater than 100×.

Therefore, increasing the voltage from 1 V rms to 100 V rms is a 40 dB increase.

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Correct answer: 6 dB

Two equal resistors in series form a simple voltage divider. With \(100\ \Omega + 100\ \Omega\), the input voltage is split equally across the two resistors. The output taken across one resistor is therefore half of the input voltage:

\[ \frac{V_{out}}{V_{in}} = \frac{1}{2} \]

Voltage attenuation in decibels is calculated as:

\[ \mathrm{Attenuation(dB)} = 20 \log_{10}\!\left(\frac{V_{out}}{V_{in}}\right) \]

Substituting:

\[ 20 \log_{10}(0.5) \approx -6\ \mathrm{dB} \]

(The negative sign indicates attenuation.)

• 3 dB corresponds to about a \(0.707\) voltage ratio, not one half.
• 50 dB would require a very large voltage reduction, not achievable with only two equal resistors.
• 100 dB represents extreme attenuation and is far beyond what this simple divider produces.

Therefore, the voltage attenuation of the network is approximately 6 dB.

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Correct answer: 20 dB

Voltage attenuation in decibels is calculated using:

\[ \mathrm{Attenuation(dB)} = 20 \log_{10}\!\left(\frac{V_{out}}{V_{in}}\right) \]

Here, the input voltage is \(V_{in} = 10\ \mathrm{V_{rms}}\) and the output voltage is \(V_{out} = 1\ \mathrm{V_{rms}}\), giving a ratio of:

\[ \frac{V_{out}}{V_{in}} = \frac{1}{10} = 0.1 \]

Substituting:

\[ 20 \log_{10}(0.1) = 20 \times (-1) = -20\ \mathrm{dB} \]

The negative sign indicates attenuation, so the magnitude of the attenuation is 20 dB.

• 6 dB corresponds to about a 2× voltage change, not a 10× change.
• 10 dB corresponds to about a 3.16× voltage change.
• 40 dB would require a 100× voltage reduction.

Therefore, reducing the voltage from 10 V rms to 1 V rms corresponds to an attenuation of 20 dB.

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