A 500 W heater is designed to operate at 200 V potential difference. If it is connected across 160 V line, the heat it will produce in 20 minutes is

Options

A. 384 kJ
B. 483 kJ
C. 843 kJ
D. 348 kJ

Answer

A

Explanation

Case-I Resistance of heater when it operates at 200 V and power, `P_(1)=500W`
`because " Power", P=V_(2)/R rArr R_(1)=V_(1)^(2)/P_(1)`
`" "R_(1)=(200)^(2)/500=(200 times 200)/500`
`" "R_(1)=80 Omega`
Case-II Now, it connected across 160 V line the heat it will produce in 20 minute is
`V_(2)=160 V`
`therefore " "P_(2)=V_(2)^(2)/R=(160)^(2)/R`
`" "(160)^(2)/80=(160 times 160)/80=2 times 160 " "(because R=80Omega)`
`P_(2)=320 W`
`therefore` Now, Heat produce is 20 minutes
`" "dQ=P_(2) times " time "=320 times 20 times 60`
`" "dQ=384kJ`