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Electric fuse is a protective device use...

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

A

300 V

B

190V

C

250 V

D

220 V

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The correct Answer is:
B

b. `I prop r^(3//2)`
`I_1/I_2 = (4r)^(3//2)/(r^(3//2) = 8/1`
`P = v^2/ R or 2 xx 10^3 = V^2/20`
or `V = 200V`
`eta = R/(R+r) xx 100 `, where R = r, efficiency will be maximum.
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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

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 . Two fuse wires of same potential , material have length ratio 1:2 and ratio of radius 4:1 . Then respective ratio of their current rating will be

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