A signal of 0.1 kw is transmitted in a cable. The attenuation of the cable is -5 dB per km and cable length is 20km. The power recieved at receiver is 10^(-x) W. The value of x is: [Gain in dB= 10log(Output Power/Input Power)]

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)))]

In a common emitter transistor amplifier, the output resistance is 500KOmega and the current gain beta=49 . If the power gain of the amplifier is 5xx10^(6) , the input resistance is

The current gain in the common emitter mode of a transistor is 10. The input impedance is 20kOmega and load of resistance is 100k Omega . The power gain is

The length of the tube of a microscope is 10 cm . The focal lengths of the objective and eye lenses are 0.5 cm and 1.0 cm . The magnifying power of the microscope is about

Electric fuse is a protective device used in series with an electric circuit or an electric appliance to save it from damage due to overheating porduced by strong current in the circuit or application. Fuse wire is generally made from an alloy of lead and tin, which has hifh resistance and low melting point. It is connected in series in an electric installation. If a circuit gets accidentally short-circuited, a large current flows, then fuse wire melts away, which causes a break in the circuit. The power through fuse (F') is equal to heat energy lost per unit area per unit time (h) (neglecting heat loses from ends of the wire). P = I^2R = h xx 2pirl [R = (rhol)/(pir^2)] where r and l are the length and radius of fuse wire, respectively. A Battery is described by its emf (E) and internal resistance (r). Efficiency of battery (eta) is defined as the ratio of the output to the input power. eta = (Output power)/(Input power) xx 100% But I= E//(R+r), input power = EI. Output power = EI - I^2r , then. eta = ((EI - I^2r)/(EI)) xx 100 = (1-(Ir)/E) xx 100 =[1 - (E/(R+r)) (r/E)] xx 100 = (R/(R+r)) xx 100 We know that output power of a source is maximum when the external resistance is equal to internal resistance, i.e., R = r . The maximum power rating of a 20.0 Omega fuse wire is 2.0 kW. This fuse wire can be connected safely to a DC source (negligible internal resistance) of

Electric fuse is a protective device used in series with an electric circuit or an electric appliance to save it from damage due to overheating porduced by strong current in the circuit or application. Fuse wire is generally made from an alloy of lead and tin, which has hifh resistance and low melting point. It is connected in series in an electric installation. If a circuit gets accidentally short-circuited, a large current flows, then fuse wire melts away, which causes a break in the circuit. The power through fuse (F') is equal to heat energy lost per unit area per unit time (h) (neglecting heat loses from ends of the wire). P = I^2R = h xx 2pirl [R = (rhol)/(pir^2)] where r and l are the length and radius of fuse wire, respectively. A Battery is described by its emf (E) and internal resistance (r). Efficiency of battery (eta) is defined as the ratio of the output to the input power. eta = (Output power)/(Input power) xx 100% But I= E//(R+r), input power = EI. Output power = EI - I^2r , then. eta = ((EI - I^2r)/(EI)) xx 10 (1-(Ir)/E) xx 100 =1 - (E/(R+r)) (r/E) xx 100 = (R/(R+r)) xx 100 We know that output power of a source is maximum when the external resistance is equal to internal resistance, i.e., R = r . Efficiency of a battery (nonideal) when delivering the maximum power is

A dry cell of emf 1.5 V and internal resistance 0.10 Omega is connected across a resistor in series with a very low resistance ammeter. When the circuit is switched on , the ammeter reading settles to a steady value of 2.0 A . (i) What is the steady rate of chemical energy consumption of the cell ? (ii) What is the steady rate of energy dissipation inside the cell ? (ii) What is the steady rate of energy dissipation inside the resistor ? (iv) What is the steady power out put of the source?

A battery is supplying power to a tape recorder by cable of resistance of 0.02 Omega . If the battery is generating 50 W power at 5 V, then the power received by the tape recorder is