
An electric heater of resistance $8\Omega $ draws a current of $15A$ from the service mains in $2hours$ . Calculate the rate at which the heat is developed in the heater.
Answer
574.5k+ views
Hint: Use Joule’s Law of heating Effect to solve this question.
Joule’s Law of heating effect gives a relationship between the current , resistance and time.
Use the formula $H = {I^2}Rt$ to solve the question.
Complete Step by step Answer
We know that when current flows across a resistive element , it produces heat.
Current= The rate of flow of charges. It is denoted by $I$. And it is given by
$I = \dfrac{q}{t}$
$ \Rightarrow q = I \times t$…………….$equation\left( 1 \right)$
Where $q = $charge
$t = $ time for which the current flows
We also know that potential Energy of the battery is used to provide energy for heat generation.
Potential=The amount of work done in bringing a unit positive charge from infinity to a point in the electric field . It is denoted by $V$
$V = \dfrac{W}{q}$
Where $W = $Work done
$q = ch\arg e$
So from the above equation we have
$W = V \times q$
Now from $equation\left( 1 \right)$ , we have
$q = I \times t$
So now the formula for work done becomes
$W = V \times I \times t$
Now we also know that according to Ohm’s Law
$V = IR$
Where $V = $ Potential Difference
$I = $ Current
$R = $ Resistance
Resistance=It is defined as the opposition to the flow of current. It’s unit is $\Omega $ .
Using the value of V in the above equation, we have
$W = IR \times I \times t$
$W = {I^2}Rt$
This work done is generated as heat.
So $W = H$
$ \Rightarrow H = {I^2}Rt$
The above equation is known as Joule’s Law of heating effect.
Which states that
$H \propto {I^2}$
$H \propto R$
$H \propto t$
Thus using the formula of heat we can solve the above question as
$H = {I^2}Rt$
Given, $I = 15A$
$R = 8\Omega $
$t = 2hours$
$t = 2 \times 60 \times 60$
$t = 7200\sec $
Putting these values in the formula of heat , we get
$H = {15^2} \times 8 \times 7200$
$H = 12960000J$
$H = 1296 \times {10^4}J$
Note:
Please keep the unit considerations in mind .The time which is given in hours has to be converted to seconds in order to get the correct question. The current has to be in Amperes and Resistance in ohms.
Joule’s Law of heating effect gives a relationship between the current , resistance and time.
Use the formula $H = {I^2}Rt$ to solve the question.
Complete Step by step Answer
We know that when current flows across a resistive element , it produces heat.
Current= The rate of flow of charges. It is denoted by $I$. And it is given by
$I = \dfrac{q}{t}$
$ \Rightarrow q = I \times t$…………….$equation\left( 1 \right)$
Where $q = $charge
$t = $ time for which the current flows
We also know that potential Energy of the battery is used to provide energy for heat generation.
Potential=The amount of work done in bringing a unit positive charge from infinity to a point in the electric field . It is denoted by $V$
$V = \dfrac{W}{q}$
Where $W = $Work done
$q = ch\arg e$
So from the above equation we have
$W = V \times q$
Now from $equation\left( 1 \right)$ , we have
$q = I \times t$
So now the formula for work done becomes
$W = V \times I \times t$
Now we also know that according to Ohm’s Law
$V = IR$
Where $V = $ Potential Difference
$I = $ Current
$R = $ Resistance
Resistance=It is defined as the opposition to the flow of current. It’s unit is $\Omega $ .
Using the value of V in the above equation, we have
$W = IR \times I \times t$
$W = {I^2}Rt$
This work done is generated as heat.
So $W = H$
$ \Rightarrow H = {I^2}Rt$
The above equation is known as Joule’s Law of heating effect.
Which states that
$H \propto {I^2}$
$H \propto R$
$H \propto t$
Thus using the formula of heat we can solve the above question as
$H = {I^2}Rt$
Given, $I = 15A$
$R = 8\Omega $
$t = 2hours$
$t = 2 \times 60 \times 60$
$t = 7200\sec $
Putting these values in the formula of heat , we get
$H = {15^2} \times 8 \times 7200$
$H = 12960000J$
$H = 1296 \times {10^4}J$
Note:
Please keep the unit considerations in mind .The time which is given in hours has to be converted to seconds in order to get the correct question. The current has to be in Amperes and Resistance in ohms.
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