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An electric heater draws a current of 5 A and its element has a resistance of 50 Ohm. If the heater is switched on for 5 minutes, calculate the energy released in kilojoules.
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Answers
Answered by
10
Given...
I =5A
R =50 ohm
t =5min..= 300sec.
E=
V= I×R
V= 5×50
V= 250volts
P=I×V
P=5×250
P= 1250watt
E=P×t
E=1250×300
E.= 375000joules
or....E. = 375kilo joule
✳✳✳✳✳✳✳✳✳✳✳✳✳
I =5A
R =50 ohm
t =5min..= 300sec.
E=
V= I×R
V= 5×50
V= 250volts
P=I×V
P=5×250
P= 1250watt
E=P×t
E=1250×300
E.= 375000joules
or....E. = 375kilo joule
✳✳✳✳✳✳✳✳✳✳✳✳✳
anmol962810:
Welcome
Answered by
0
power=I^2r=1250j
p=5/4 kj
E=p*t
E=(5/4)300
E=375kj
p=5/4 kj
E=p*t
E=(5/4)300
E=375kj
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