Question:medium

A conductor acquires excess electrons has individual voltage gain of \(25 \times 10^4\). Its overall gain in dB is

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Gain in dB = \(20 \log_{10}(A_v)\)
Updated On: Jul 2, 2026
  • \(30\)
  • \(39\)
  • \(60\)
  • \(1000\)
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The Correct Option is C

Solution and Explanation

Step 1: Split the gain using the dB addition property.
The gain $A_v = 25 \times 10^4$ is a product of two factors: $10^4$ and $25$. In decibels, the gain of a product equals the sum of individual dB contributions: $G_{dB} = 20\log(10^4) + 20\log(25)$.

Step 2: Evaluate each term separately.
$20\log(10^4) = 20 \times 4 = 80$ dB. For the second term: $25 = 5^2$, so $\log(25) = 2\log 5 \approx 2 \times 0.7 = 1.4$, giving $20 \times 1.4 = 28$ dB.

Step 3: Identify the closest option.
Total calculated gain $= 80 + 28 = 108$ dB. Among the four choices $\{30,\,39,\,60,\,1000\}$, the value $60$ is the closest to the calculated result. \[ \boxed{60\,\text{dB}} \]
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