Question

An electric heater rated 220V and 2.2KW works for 3 hours. Find the energy consumed and the current drawn by it. [Ans: 2.376 × 107J.​

Answer 1

Answer:

Current: $10\; {\rm A}$.

Energy consumed: $2.376 \times 10^{7}\; {\rm J}$.

Explanation:

Convert all values to standard units:

$P = 2.2\; {\rm kW} = 2.2 \times 10^{3}\; {\rm W}$.

$V = 220\; {\rm V}$.

$t = 3\; {\rm hr} = 3 \times 3600\; {\rm s} = 10800\; {\rm s}$.

The power consumption $P$ of an electric appliance is the product of:

• the voltage $V$ of this appliance, and
• the current $A$ going through this appliance.

In other words:

$P = V\, A$.

Rearrange this equation to find current $A$ in terms of power rating $P$ and voltage $V$:

$\begin{aligned}A &= \frac{P}{V}\end{aligned}$.

In this question, it is given that $P = 2.2 \times 10^{3}\; {\rm W}$ while $V = 220\; {\rm V}$. Thus, the current going through this appliance would be:

$\begin{aligned}A &= \frac{P}{V} \\ &= \frac{2.2 \times 10^{3}\; {\rm W}}{220\; {\rm V}} \\ &= 10\; {\rm A}\end{aligned}$.

The power rating of this appliance gives the amount of energy that this appliance consumes per unit time. The energy that this $P = 2.2 \times 10^{3}\; {\rm W}$ appliance would consume in $t = 10800\; {\rm s}$ would be:

$\begin{aligned}E &= t\, P \\ &= (10800\; {\rm s}) \times (2.2 \times 10^{3}\; {\rm W}) \\ &= (10800\; {\rm s}) \times (2.2 \times 10^{3}\; {\rm J \cdot s^{-1}}) \\ &= 2.376 \times 10^{7}\; {\rm J}\end{aligned}$.