Question

A geyser is rated $1500W,250V.$ This geyser is connected to $250V$ mains. Calculate:
(A) The current drawn,
(B) The energy consumed in $50\,hours$ and
(C) The cost of energy consumed at ₹ $4.20{\text{ per kWh}}$

Answer 1

Hint
We are provided with the power and voltage supply of a geyser. We have to find the current drawn by the geyser, the energy consumed by it in a given time and the cost of the energy consumed. To find the answer use the power formula and electric energy formula.

Complete step by step answer
When a current I flows through a conductor of resistance R in time t, then the quantity of charge flowing in the conductor is given by
$\Rightarrow q = It{\text{ }} \to {\text{(1)}}$
Where, q is the charge flowing through the conductor, I is the current through the conductor, t is the time.
And, When the charge q, flows between two points having a potential difference V, then the work done in moving the charge is given by
$\Rightarrow W\; = V \times q{\text{ }} \to {\text{(2)}}$
We know that
$q = It.$
So,
$\Rightarrow W = V \times It{\text{ }} \to {\text{(3)}}$
Where, W is the work done, V is the voltage difference.
Now, we can define an electric power
Electric power is the rate of doing electrical work when voltage is applied. Power is given by the ratio of work done by time
$\Rightarrow P = \dfrac{W}{t}{\text{ }} \to {\text{(4)}}$
Where,
$W = V \times It$
So,
$\Rightarrow P = \dfrac{{VIt}}{t}$
$\Rightarrow P = VI{\text{ }} \to {\text{(5)}}$
Now, power is the product of potential difference and current strength.
We know that ohm’s law equation
$V = IR$
If we substitute this ohm’s equation in the power formula we get
$\Rightarrow P = {I^2}R{\text{ }} \to {\text{(6)}}$
The unit of power is Watt (W).
Electric energy is defined as the capacity to do work. The electric energy is given by the product of power and time.
Electric energy is given by the formula
$E = Pt{\text{ }} \to {\text{(7)}}$
Where, P is the power, t is the time.
Its unit is joule. The electrical energy is measured by watt hour in practice.
We can convert watt to joule by
$\Rightarrow 1{\text{ }}kWh{\text{ }} = {\text{ }}1000{\text{ }}Wh{\text{ }} = {\text{ }}1000{\text{ }} \times {\text{ }}3600{\text{ }}J{\text{ }} = {\text{ }}36{\text{ }} \times {\text{ }}{10^5}{\text{ }}J$
Given,
The electric power of the geysers, $P = 1500W$
The voltage applied, $V = 250V$
-We have to find the current drawn by the geyser,
From the equation-5
$\Rightarrow P = VI$
$\Rightarrow I = \dfrac{P}{V}$
Substitute the given values,
$\Rightarrow I = \dfrac{{1500}}{{250}}$
$\Rightarrow I = 6A$
-To find the energy consumed in $50\,hours$
Given,
$t = 50hrs$
E=?, From equation-7
$\Rightarrow E = Pt$
$\Rightarrow E = 1500 \times 50$
$\Rightarrow E = 75000\,Wh$
$\Rightarrow E = 75 \times {10^3}\,Wh$
$\Rightarrow E = 75\,k\,Wh$
-The cost of energy consumed at ₹ $4.20{\text{ per kWh}}$
Given, The cost of energy consumed for 1 KWh is $4.20$
The cost of energy consumed for $75\,kWh$ is given by
$\Rightarrow 75\, \times 4.20$
The cost of energy consumed for $75kWh$ is ₹ $315$ .

Note
In this problem, in the second section they have asked us to find the energy consumed in $50\,hours$ , so no need to change the hours into seconds we have to use as it is. Similarly, in the third division the cost of energy is given for one kilowatt so, we have to use $75kwh$ not $75000Wh.$