
An electric heater consumes $1.1kW$ power when $230V$the voltage is applied to it. How much current will be flowing through it?
$\left( a \right)$ $1.1A$
$\left( b \right)$ $2.2A$
$\left( c \right)$ $4A$
$\left( d \right)$ $5A$
Answer
180k+ views
.Hint. In this question, we have an electric heater which consumes power, and the voltage is applied to it. We have to find the current flowing through it. So by using the formula of power, $P = V/I$, we will be able to find the current.
.Formula used.
Power,
$P = \dfrac{V}{I}$
Here,
$P$, will be the power
$V$, will be the voltages
$I$, will be the current.
.Complete Step By Step Solution. First of all we will convert the power given to the watt.
So,
Power$ = 1.1kwh = 1.1 \times 1000$
Then the final power will be
$P = 1100w$
We know that,
$P = \dfrac{V}{I}$
From there the current will be
$ \Rightarrow I = P/V$
Now putting the values, we get
$ \Rightarrow I = \dfrac{{1100}}{{220}}$
$ \Rightarrow I = \dfrac{{10}}{2}$
And on solving the above, we get
$ \Rightarrow 5A$
Therefore, $5A$ the current will be following through it.
Hence option $D$will be the correct one.
Additional information: Electric force is generally provided to organizations and homes (as homegrown mains power) by the electric force industry through an electric force network. The electric force is normally sold by the kilowatt-hour which is the result of the force in kilowatts duplicated by running time in hours. Electric utilities measure power utilizing a power meter, which keeps a running complete of the electric energy conveyed to a client.
.Note. Electrical Power is the rate at which electrical energy is used. For example, consider 1 Joule of electrical energy. If one uses this in one second then the power is one Watt. If one uses this same energy in a millisecond the power is one $kW$ and if used in a microsecond then the power is 1$mW$. So the same electrical energy yields different electrical power depending upon the rate at which it is used.
This definition holds for all forms of power and is the basis of Newton’s first and second laws. No power, no change in energy, no change in state or potential.
.Formula used.
Power,
$P = \dfrac{V}{I}$
Here,
$P$, will be the power
$V$, will be the voltages
$I$, will be the current.
.Complete Step By Step Solution. First of all we will convert the power given to the watt.
So,
Power$ = 1.1kwh = 1.1 \times 1000$
Then the final power will be
$P = 1100w$
We know that,
$P = \dfrac{V}{I}$
From there the current will be
$ \Rightarrow I = P/V$
Now putting the values, we get
$ \Rightarrow I = \dfrac{{1100}}{{220}}$
$ \Rightarrow I = \dfrac{{10}}{2}$
And on solving the above, we get
$ \Rightarrow 5A$
Therefore, $5A$ the current will be following through it.
Hence option $D$will be the correct one.
Additional information: Electric force is generally provided to organizations and homes (as homegrown mains power) by the electric force industry through an electric force network. The electric force is normally sold by the kilowatt-hour which is the result of the force in kilowatts duplicated by running time in hours. Electric utilities measure power utilizing a power meter, which keeps a running complete of the electric energy conveyed to a client.
.Note. Electrical Power is the rate at which electrical energy is used. For example, consider 1 Joule of electrical energy. If one uses this in one second then the power is one Watt. If one uses this same energy in a millisecond the power is one $kW$ and if used in a microsecond then the power is 1$mW$. So the same electrical energy yields different electrical power depending upon the rate at which it is used.
This definition holds for all forms of power and is the basis of Newton’s first and second laws. No power, no change in energy, no change in state or potential.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2025-26 Mock Test: Dual Nature of Matter & Radiation

JEE Main Mock Test 2025-26: Dual Nature of Matter & Radiation

Young's Double Slit Experiment Step by Step Derivation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

What is Hybridisation in Chemistry?

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Wheatstone Bridge for JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation
