A heater having 400 resistance is connected in 200volt supply.lf the heater is worked for 10 min.calcuate the quatity of heat produced
Answers
Answer:
As given in question :
P= 1000 watt ( assumed that temprature remain constant)
V= 250 V so, I ( current) = P÷V = 1000÷250 = 4 Amperes.
And power dissipation 1000 watt = I^2R that is 4×4×R or R = 1000÷16 = 62.5 ohms (at constant temprature).
If we consider that temprature is constant,
The power dissipation for applied 200 V will be V^2 ÷R or 200×200/ 62.5 = 40000÷62.5 = 640 watt. It's a theoretical explanation.
But Practically, when we usee the heater the temprature of coiled wire/ material increases which in turn increases the resistance of the coil (used material) which results increase in power dissipation. If this increase of heat reaches nearer to the melting point of the material used, the resistance of material goes too high which results break with spark in the coiled wire.
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