Question

A geyser is rated $1500\,\,W$, $250\,\,V$. This geyser is connected to $250\,\,V$ mains. The energy consumed in $50$ hours is:
(A) $75\,\,KWh$.
(B) $50\,\,KWh$.
(C) $1500\,\,KWh$.
(D) $250\,\,KWh$.

Answer 1

Hint
Energy consumption refers to all the energy used to perform an action, manufacture something or simply inhabit a building. Let's look at a few examples: In a factory, total energy consumption can be measured by looking at how much energy a production process consumes.
The energy consumed by the geyser is given by the formula;
$\Rightarrow W = P \times t$
Where, $W$ denotes the energy consumed by the geyser, $P$ denotes the power rated for the geyser, $t$ denotes the time taken.

Complete step by step answer
The data given in the problem are;
Power rated for the geyser is, $P = 1500\,\,W$.
The geyser connected to the main is, $v = 250\,\,V$.
Time taken by the geyser is, $t = 50\,\,h$.
The power of the geyser is given as;
$\Rightarrow P = v \times I$
Where, $P$ denotes the power of the geyser, $v$ denotes the voltage source in which the geyser is connected, $I$ denotes the current required by the geyser.
Substitute the values of voltage and current on the above formula;
$\Rightarrow 1500 = 250 \times I$
$\Rightarrow I = \dfrac{{1500}}{{250}}$
$\Rightarrow I = 6\,\,A$
The energy consumed by the geyser in $KWh$ is given by the formula;
$\Rightarrow W = P \times t$
Substitute the values of power of the geyser and the time taken by it to run;
$\Rightarrow W = 1500\,\,W \times 50\,\,h$
$\Rightarrow W = 1500 \times 50$
$\Rightarrow W = 75000\,\,Wh$
Since we only need it in the form of $KWh$;
$\Rightarrow W = 75\,\,KWh$.
Therefore, the energy consumed in $50\,\,\,h$ is calculated as $W = 75\,\,KWh$.
Hence, the option (A) $W = 75\,\,KWh$ is the correct answer.

Note
In electrical engineering, power consumption refers to the electrical energy per unit time, supplied to operate something, such as a home appliance. Power consumption is usually measured in units of watts $\left( W \right)$ or kilowatts $\left( {KW} \right)$.