Question

An electric heater of power 1000 Watt is joined to electric mains of 250 Volts Calculate:
(i) Electric current flowing through the heater
(ii) Resistance of the heater wire
(iii) Thermal energy produced per minute by heater.
(iv) The electric energy spent in KiloWatt-Hour by heater in $2$ hour

Answer 1

Hint: The question is based on Ohm’s law, electrical power calculations and the definition of $1$ unit as calculated in the standard electrical bill in India. Also the convention of $1$ unit in the terms of joules is required.

Formula used:
1. ${{P}}\;{{ = }}\;{{VI}}$, Here P is the power and V and I are the voltage and current respectively.
2. $V = IR$, here V is the voltage and I and R are the current and resistance respectively.
3. ${{E}}\;{{ = P \times t}}$, here E is the energy and P and t are the power and time respectively.${{1}}\;{{unit = }}\;{{1}}\;{{kwh}}\;{{ = 3}}{{.6 \times 1}}{{{0}}^{{6}}}{{J}}$

Complete step by step solution:
Given, here power
 $
  {{P}}\;{{ = 1000}}\;W \\
    \\
 $
$V = 250V$

(i) We know that, Power
${{P}}\;{{ = V \times I}}$
$ \Rightarrow {{I}}\;{{ = }}\dfrac{P}{V}$
Putting the values, we get
${{I}}\;{{ = }}\dfrac{{{{1000}}}}{{{{250}}}}$
$ \Rightarrow I = 4A$
So, Electric current flowing through the heather i.e. \[{{I}}\;{{ = }}\;{{4A}}\]

(ii) We have to calculate resistance of the heater wire by Ohm’s law, ${{R}}\;{{ = }}\;\dfrac{{{V}}}{{{I}}}$
Where, R= resistance
V= Voltage
I= current
So, putting the values we get
${{R}}\;{{ = }}\dfrac{{{V}}}{{{{I}}\;}}\;{{ = }}\;\dfrac{{{{250}}}}{{{4}}}\;\;{{ = }}\;{{62}}{{.5\Omega }}$
So, Resistance of the heater wire is $62.5\Omega $

(iii) Now we have to calculate thermal energy produced per minute by the heater.
We know,
Energy can be given by,
${{E = }}\;{{P \times t}}$
Given, Power ${{ = }}\;{{1000W}}$
Time $ = \;1\;\min $
$ \Rightarrow t = 60\;\sec $
We have to convert min into seconds because we have to calculate energy in Joules $\left( {{{S}}{{.I}}{{.}}\;{{Unit}}} \right)$
So, putting the values, we get
${{E = }}\;{{P \times t}}\;{{ = 1000 \times 60}}\;{{ = }}\;{{60000J}}{{.}}$
So, thermal energy produced per minute by the heater is ${{60,000J}}{{.}}$

(iv) Now, we have to calculate-
Electrical energy spent in kilowatt-hour by heater in $2$ hours-
We know that
Energy consumed ${{ = }}\;P \times t$
Power (given) ${{ = }}\;{{1000W}}$
${{ = 1KW}}$
Time (given) ${{ = }}\;{{2}}\;{{hours}}$
So, putting the values we get
Energy consumed${{ = }}\;{{P \times t}}\;{{ = 1}}\;{{KW \times 2}}\;{{hours}}$
\[{{ = }}\;{{2KWh}}\]

Additional Information:
This question has many concepts which have to be remembered by the students.
(i) The first concept is Ohm’s Law. It states that the voltage$\left( {{V}} \right)$ across a resistor is always equals to current in the resistor $\left( {{I}} \right)$ times resistance of the resistor $\left( {{R}} \right)$.
(ii) The second concept which is involved in this question is the concept of electrical power, ${{P = }}\;{{V \times I}}\;{{ = }}\;{{{I}}^{{2}}}{{R}}\;{{ = }}\dfrac{{{{{V}}^{{2}}}}}{{{R}}}$
(iii) The concept of unit of electricity consumed in relation with energy in Joules.
${{1KWh}}\;{{ = }}\;{{1Unit}}\;{{ = }}\;{{3}}{{.6 \times 1}}{{{0}}^{{6}}}{{J}}$.

Note:Ohm’s Law states that for any resistor the potential difference across it, ${{V,}}$ is equal to the product of current in the resistor $\left( {{I}} \right)$, times the resistance of the resistor $\left( {{R}} \right)$ provided the temperature of the resistance remains constant.