A signal of `0.1kW` 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^(-x)W`. The value of `x` is `-----` [Gain in `dB=10log_(10)((P_(0))/(P_(i)))`]

A 1 KW signal is transmitted using a communication channel which provides attenuatiom at the rate of - 2 dB per km . If the communication channel has a total length of 5 km , the power of the signal received is [ gain in dB = 10 log ((P_(0))/(P_(i)))]

A 1 KW signal is transmitted using a communication channel which provides attenuatiom at the rate of - 2 dB per km . If the communication channel has a total length of 5 km , the power of the signal received is [ gain in dB = 10 log ((P_(0))/(P_(i)))]

A CE amplifier, has a power gain of 40 dB. The input resistance and output load resistance are 500Omega and 2kOmega respectively. Find the value of common - emitter current gain. ["take gain in dB "=10log((P_("out"))/(P_("in")))]

If the distance between the points (4,p) and (1,0) is 5 , then the value of p is

Figure (EP) shows a communication system. What is the output power when input signals is of 1.01 mW? [ gain in dB = 10 log_10(P_0//P_t)] .

Figure (EP) shows a communication system. What is the output power when input signals is of 1.01 mW? [ gain in dB = 10 log_10(P_0//P_t)] .

If log_(10) 4 = 0.6020 , find the value of : (i) log_(10) 8 (ii) log_(10) 2.5