A) \[1.3\,dB\]
B) \[3.1\,dB\]
C) \[13\,dB\]
D) \[30.1\,dB\]
Correct Answer: C
Solution :
As, \[p\propto I\] Increase in power in dB \[S{{L}_{2}}-S{{L}_{1}}=10\log \left( \frac{{{I}_{2}}}{{{I}_{1}}} \right)=10\log \left( \frac{{{p}_{2}}}{{{p}_{1}}} \right)\] \[=10\log \left( \frac{400}{20} \right)\] \[-10\log 20=12dB\]You need to login to perform this action.
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