
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
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.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.
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