If the gain of an amplifier is 18 dB, what is the new gain if the power is reduced by half?

If the gain of an amplifier is 18 dB, what is the new gain if the power is reduced by half?

• A. 12 dB
• B. 15 dB
• C. 10 dB
• D. 9 dB

Answer: (B)

To determine the new gain when the power is reduced by half, it's essential to understand the relationship between gain in decibels (dB) and power. Gain in dB can be calculated using the formula: \[ \text{Gain (dB)} = 10 \log_{10} \left( \frac{P_2}{P_1} \right) \] where \( P_2 \) is the output power and \( P_1 \) is the input power. When the power is reduced by half, \( P_2 = \frac{P_1}{2} \). Plugging this into the formula gives: \[ \text{Gain (dB)} = 10 \log_{10} \left( \frac{1}{2} \right) = 10 \log_{10}(0.5) \] Using the logarithmic property, we find that: \[ \log_{10}(0.5) \approx -0.301 \] Thus, \[ \text{Gain (dB)} = 10 \times -0.301 \approx -3.01 \, \text{dB} \] When you reduce the original gain,

To determine the new gain when the power is reduced by half, it's essential to understand the relationship between gain in decibels (dB) and power. Gain in dB can be calculated using the formula:

[ \text{Gain (dB)} = 10 \log_{10} \left( \frac{P_2}{P_1} \right) ]

where ( P_2 ) is the output power and ( P_1 ) is the input power. When the power is reduced by half, ( P_2 = \frac{P_1}{2} ). Plugging this into the formula gives:

[ \text{Gain (dB)} = 10 \log_{10} \left( \frac{1}{2} \right) = 10 \log_{10}(0.5) ]

Using the logarithmic property, we find that:

[ \log_{10}(0.5) \approx -0.301 ]

Thus,

[ \text{Gain (dB)} = 10 \times -0.301 \approx -3.01 , \text{dB} ]

When you reduce the original gain,