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Question
An electrical appliance is rated at 1000 KVA, 220V. If the appliance is operated for 2 hours, calculate the energy consumed by the appliance in: (i) kWh (ii) joule.
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Solution
Given: V = 220 volt, 1000 KVA
Time = 2 hours
(i) In kWh Energy consumed = Pt = 2000 kWh
(ii) In joule
We know, 1kWh = 3.6 × 106 J
So, 2000 kWh = `(2000 xx 3.6 xx 10^6)/1`
= 7.2 × 109 Joule
= 7.2 × 109 J
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