A) \[13dB\]
B) \[10dB\]
C) \[20dB\]
D) \[800dB\]
Correct Answer: A
Solution :
\[P\propto I\] \[\therefore \]\[{{L}_{1}}=10{{\log }_{10}}\left( \frac{{{I}_{1}}}{{{I}_{0}}} \right)\]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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