Q.

A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is –5 dB per km and cable length is 20 km. The power received at receiver is  10−xW. The value of ‘x’ is …… Given in dB=10log10⁡P0Pi

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answer is 8.

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Detailed Solution

If sound level decreases by 5dB every km, then sound level decreased in 20km=100dBβo−βi=10log10⁡P0Pi⇒−100=10log10⁡P0Pi⇒100=10log10⁡PiP0⇒PiP0=1010⇒P0=Pi1010=0.1×1031010=10−8W So, x=8
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