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Question
An electrical appliance having a resistance of 200 Ω is operated at 200 V. Calculate the energy consumed by the appliance in 5 minutes in kWh.
Solution
Given resistance, R = 200 Ω
Voltage, V = 200 V
time, t = 5 min
= 5 × 60 sec
= 300 sec
As, Energy, E = `("V"^2"t")/"R"`
In kWh
As 1 kWh = 3.6 × 106 J
1 J = `1/(3.6 xx 10^6)` kWh
60000 J = `60000/(3.6 xx 10^6)` = 0.0167 kWh
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