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: 100

Voltage gain in decibels is given by:

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

Given:

\[ 40 = 20 \log_{10}\left(\frac{V_{\text{out}}}{V_{\text{in}}}\right) \]

Dividing both sides by 20:

\[ 2 = \log_{10}\left(\frac{V_{\text{out}}}{V_{\text{in}}}\right) \]

Taking the antilog:

\[ \frac{V_{\text{out}}}{V_{\text{in}}} = 10^2 = 100 \]

• 20 corresponds to about 26 dB.
• 40 corresponds to about 32 dB.
• 400 corresponds to about 52 dB.

Therefore, the voltage ratio is 100.

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Correct answer: D — 100

A gain of 20 dB expressed as a power ratio uses the decibel power formula. Every 10 dB represents a factor of 10 in power, so 20 dB represents 10 × 10 = 100. More formally:

\[ \text{Gain (dB)} = 10 \log_{10}\!\left(\frac{P_{\text{out}}}{P_{\text{in}}}\right) \]

Rearranging to find the power ratio:

\[ \frac{P_{\text{out}}}{P_{\text{in}}} = 10^{\,\text{Gain(dB)}/10} \]

Substituting 20 dB:

\[ \frac{P_{\text{out}}}{P_{\text{in}}} = 10^{20/10} = 10^{2} = 100 \]

• A — 10: This corresponds to a gain of 10 dB, not 20 dB.
• B — 20: 20 is the gain value in decibels itself, not the power ratio. Confusing the dB figure for the linear ratio is a common error.
• C — 40: This has no standard dB equivalent; it may result from incorrectly doubling the 20 dB figure linearly.

Therefore, a 20 dB power gain means the output power is exactly 100 times the input power.

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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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Correct answer: 6 dB

Voltage attenuation in decibels is given by:

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

Given:

• \(V_{\text{in}} = 10\ \mathrm{V_{rms}}\)
• \(V_{\text{out}} = 5\ \mathrm{V_{rms}}\)

Substituting:

\[ \text{Attenuation} = 20 \log_{10}\left(\frac{10}{5}\right) = 20 \log_{10}(2) \approx 20 \times 0.301 \approx 6\ \mathrm{dB} \]

• 10 dB would correspond to a larger voltage ratio.
• 20 dB represents a voltage ratio of 10:1.
• 40 dB represents a voltage ratio of 100:1.

Therefore, the attenuation is 6 dB.

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Correct answer: 50 dB

When amplifiers are connected in cascade, their gains in decibels add directly:

\[ G_{\text{total}} = G_1 + G_2 \]

Given:

• \(G_1 = 10\ \mathrm{dB}\)
• \(G_2 = 40\ \mathrm{dB}\)

So:

\[ G_{\text{total}} = 10 + 40 = 50\ \mathrm{dB} \]

• 8 dB and 30 dB are incorrect sums.
• 400 dB would result from incorrectly multiplying dB values.

Therefore, the total gain is 50 dB.

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Correct answer: B — 10 dB

When stages are connected in cascade, their gains and losses in decibels are simply added algebraically. An attenuator introduces a negative gain (a loss), so combining a +20 dB amplifier with a −10 dB attenuator gives an overall gain of:

\[ G_{\text{total}} = G_{\text{amplifier}} + G_{\text{attenuator}} \]

\[ G_{\text{total}} = 20\ \text{dB} + (-10\ \text{dB}) = 10\ \text{dB} \]

This is one of the key advantages of the decibel scale — cascaded stages can be combined by simple addition rather than multiplication of power ratios.

• A — 8 dB: Incorrect; there is no basis for subtracting 12 dB from 20 dB in this scenario.
• C — −10 dB: Incorrect; this ignores the amplifier's 20 dB gain entirely and only states the attenuator value.
• D — −200 dB: Incorrect; this appears to be a mistaken multiplication of the two values (20 × −10), which is not how decibels are combined in cascade.

Therefore, the gain of the combination is 10 dB, found by algebraically adding the amplifier gain and attenuator loss in decibels.

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Correct answer: 15 dB

When amplifier stages are cascaded, gains in decibels add directly.

Given:

• 3 stages
• each provides 5 dB

Total gain:

\[ G_{\text{total}} = 5 + 5 + 5 = 15\ \mathrm{dB} \]

• 10 dB would be only two stages.
• 25 dB is incorrect addition.
• 125 dB would be multiplying, which is wrong for dB.

Therefore, the total amplification is 15 dB.

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