A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is -5dB per km and cable length is 20km. The power received at receiver is 10^-^xW. The value of x is…………………

[Given in dB = 10 log_{10 ^{$(\frac{P_{0}}{P_{i}})]$}}

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5 months ago

Sound level is given in dB = 10 log $\frac{_{}}{_{}} \frac{}{_{}}$ $(\frac{P_{0}}{P_{i}}) o r 10 l o g (\frac{l}{l_{0}})$

As sound level decreases 5 dB every km,

So, in 20 km sound level will decrease by 20 * 5 = 100 dB.

$Δ β = β_{2} - β_{1} = 10 l o g (\frac{l_{2}}{l_{1}})$

$- 100 = 10 l o g (\frac{l_{2}}{l_{1}})$

$1 0^{- 10} = \frac{l_{2}}{l_{1}} \Rightarrow l_{2} = 1 0^{- 10} l_{1}$

$P_{2} = 1 0^{- 10} P_{1}$

$P_{2} = 1 0^{- 10} * (0.1 * 1 0^{3}) \Rightarrow P_{2} = 1 0^{- 8} W = 1 0^{- x} W$

$∴$ x = 8

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A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is -5dB per km and cable length is 20km. The power received at receiver is 10-xW. The value of x is………………… [Given in dB = 10 log10 ( P 0 P i ) ]

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A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is -5dB per km and cable length is 20km. The power received at receiver is 10^-^xW. The value of x is…………………

[Given in dB = 10 log_{10 ^{$(\frac{P_{0}}{P_{i}})]$}}