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Assertion: Fuse wire must have high resi...

Assertion: Fuse wire must have high resistance and low melting point.
Reason: Fuse is used for small current flow only.

A

If both Assertion `&` Reason are True `&` the Reason is a correct explanantion of the Assertion

B

If both Assertion `&` Reason True but Reason is not a correct explanation of the Assertion

C

If Assertion is True but the Reason is False

D

If both Assertion `&` Reason are False.

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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 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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