A) 13 dB
B) 10 dB
C) 20 dB
D) 800 dB
Correct Answer: A
Solution :
\[P\propto I\] \[{{L}_{1}}=10{{\log }_{10}}\left( \frac{{{I}_{1}}}{{{I}_{0}}} \right)\] and \[{{L}_{2}}=10{{\log }_{10}}\left( \frac{{{I}_{2}}}{{{I}_{0}}} \right)\] So \[{{L}_{2}}-{{L}_{1}}=10{{\log }_{10}}\left( \frac{{{I}_{2}}}{{{I}_{1}}} \right)\] = \[10{{\log }_{10}}\left( \frac{{{P}_{2}}}{{{P}_{1}}} \right)\]= \[10{{\log }_{10}}\left( \frac{400}{20} \right)\]= \[10{{\log }_{10}}20\] =\[10\log (2\times 10)\] = \[10(0.301+1)=13dB\]
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