Question

An electrical appliance having a resistance of 200 ohm is operated at 200 V. Calculate the energy consumed by the appliance in 5 minutes ​
(a) in joule, ​
(b) in kWh

Answer 1

(a) Energy consumed by electric appliance in joules:
Given,
Resistance, 𝑅 = 200 𝑜ℎ𝑚
Voltage, 𝑉 = 200 𝑉
Time, 𝑡 = 5 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 = $5\times 60sec$ = 300 sec
The expression for the electrical energy (E)
spent in flow of current through an electrical
appliance in terms of potential (V), resistance
(R), and time (t) is given by,
𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 , $E=\frac{V^{2}t}{R}$
$\therefore E=\frac{{200}^2\times 300}{200}=60000Joule$.
Energy consumed by the electrical appliance is 60000 Joule.
(b) Energy consumed by electric appliance
in kWh:
Since $3.6✕{10}^{6}Joule=1kWh$
$60000J=\frac{1kwh}{3.6✕{10}^{6}Joule}\times 60000J$
= 0.01667 kWh
Energy consumed by the electrical appliance is 0.01667 kWh.