Question

An electrical appliance having a resistance of 200Ω is operated at 200 V. Calculate the energy consumed by the appliance in 5 minutes (i) in joules (ii) in kWh.

Answer 1

Given,
Resistance, R = 200 Ω

Voltage, v = 200 v
Time, t = 5 min = 5 × 60 sec = 300 sec
As, Energy, E = `(V^2t)/R`

(i) in joules 

                  E  = `((200)^2 xx 300) /200`

                  E  = 60000 J 

(ii) In kwh
As 1 kwh = 3.6 × `10^6`J
1 J = ` 1/(3.6xx10^6)` kWh 

60000 J =  `60000/(3.6xx10^6)` = 0.0167 kwh