JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Semiconductors

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

1 0^{- x} W

. The value of x is ____
[ Given in dB

= 10 lo g_{10} (\frac{P _{0}}{P _{i}})

]

Answer: 8

Solution

Over

20

km the cable attenuates the signal by

5 \times 20 = 100

dB, so

- 100 = 10 lo g_{10} \frac{P _{0}}{P _{i}} \Rightarrow \frac{P _{0}}{P _{i}} = 1 0^{- 10}

The input power is

P_{i} = 0.1 kW = 1 0^{2} W

, so the power delivered at the receiver is

P_{0} = 1 0^{2} \times 1 0^{- 10} = 1 0^{- 8} W

Comparing with

1 0^{- x}

W gives

x = 8

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Integer answer