Question

An electric iron of resistance 80 Ω is operated at 200 V for two hours. Find the electrical energy consumed.

Answer 1

Hey.

Here is the answer.

As electrical energy ,

$h = ( \frac{ {v}^{2} }{r} ) \times t$

where v = 200 V , r = 80 ohm & t = 2 hr=7200s

So,

$h = ( \frac{ {200}^{2} }{80} ) \times 7200 \\ or \: \: h \: = ( \frac{40000}{80} ) \times 7200 \\ or \: \: h = \: 500 \times 7200 \\ or \: h \: = 3600000 \: joules \\ or \: h \: = \: 3600 \: kilojoules$

As 1000 W = 1000 J/sec

1000 Wh = 1000 Jh/sec

1 kWh  = 1000 Jh/sec

1 kWh = 1000 J *3600sec/sec

Also, as 1 kWh = 3600000 J

So, energy consumed = 1 kWh.

Thanks.