Question

An electrical appliance having a resistance of $200\Omega$ is operated at $200V$. What is the energy consumed by the appliance in 5 minutes in joules ?

• A. $6000J$
• B. $40000J$
• C. $60000J$
• D. $240000J$

Answer 1

Answer: (C)

$60000J$

Energy ($power\times time$) is measured in Joules and by including time (t) in the power formulae, the energy dissipated by a component or circuit can be calculated.

Energy dissipated = Pt or $VI\times tasP=VI$.

It can also be written as $\frac{V^2}{R}\times tasI=\frac{V}{R}$.

Hence, the expression for Power in terms of voltage is given as $P=\frac{V^{2}t}{R}Joules$.

It is given that resistance is 200 ohms and voltage is 200 V. The time taken is 5 minutes, that is $5\times 60=300seconds$.

Therefore, $P=\frac{{200}^2}{200}\times 300=60,000Joules$.

Hence, the energy consumed by the electrical appliance is 60,000 J.