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

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

Energy consumed:
`2.376 * 10^(7)\; {\rm J}`.

Step-by-step explanation:

Convert all values to standard units:

`P = 2.2\; {\rm kW} = 2.2 * 10^(3)\; {\rm W}`.

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

`t = 3\; {\rm hr} = 3 * 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 &= (P)/(V)\end{aligned}`.

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

`\begin{aligned}A &= (P)/(V) \\ &= \frac{2.2 * 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 * 10^(3)\; {\rm W}` appliance would consume in
`t = 10800\; {\rm s}` would be:

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