Electric Fuse!

Asked by chuckles | 9th Sep, 2010, 06:43: AM

Expert Answer:

Power of the generator, P = 1.5 kW = 1500 Watts
Operating Voltage,        V = 220 V
 
P = VI
therefore,
The heater will generate a current I = P/V = 1500/220 = 6.8 Amperes.
 
Now the fuse put in the heater is of 5A. That is, if a current of magnitude greater than 5A passes through it, it will melt. Thus in the given heater where 6.8 A current is being generated, the fuse will melt - breaking the circuit even in the normal working of the heater. Thus, the heater will not function.
 
Change: Put a fuse of rating greater that 6.8 A.

Answered by  | 9th Sep, 2010, 07:50: PM

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