Electronics and Communication Engineering - Measurements and Instrumentation - Discussion
Discussion Forum : Measurements and Instrumentation - Section 1 (Q.No. 6)
6.
If reference sound pressure P0 is 2 x 10-5 N/m2, a sound pressure of 90 dB is equal to
Answer: Option
Explanation:
.
Discussion:
9 comments Page 1 of 1.
RAJASHEKAR REDDY said:
2 years ago
I understood it now, thanks all for explaining.
Manasa said:
7 years ago
For voltages and currents we will take 10log but for power calculations in dB form we will take it as 20log.
Dnyanesh said:
8 years ago
Excellent explanation. Thank you.
Anomi said:
8 years ago
10 log or 20 log, can somebody explain it?
Chandan kumar said:
9 years ago
Why we put here 20log?
Please derive it.
Please derive it.
Hal HJ said:
9 years ago
Why we put here 20log? What's the reason?
Dhaivat said:
9 years ago
Pressure (in dB) = 20log(P/P0)
Putting the values;
90 = 20log(P/(2 * 10^-5)).
90/20 = log(P/(2 * 10^-5)).
4.5 = log(P/(2 * 10^-5)).
10^4.5 = P/(2 * 10^-5).
Calculating, we get P = .632.
Putting the values;
90 = 20log(P/(2 * 10^-5)).
90/20 = log(P/(2 * 10^-5)).
4.5 = log(P/(2 * 10^-5)).
10^4.5 = P/(2 * 10^-5).
Calculating, we get P = .632.
Sri said:
9 years ago
I want a formula for solving the solution. If anybody knows, please provide it.
Xai said:
9 years ago
Explain the solution.
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