
An electric iron is connected to the mains power supply of 220V. When the electric iron is adjusted at ‘minimum heating’ it consumes a power of 360 W but at ‘maximum heating’ it takes a power of 840 W. Calculate the current and resistance in each case.
Answer
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Hint: We can solve this question very easily if we know the relation between power, current and voltage. First of all by using the relation we need to find the current respectively for minimum and maximum heating conditions. Then we need to find their respective values of resistances for the minimum and maximum heating condition. After this we can finally conclude with the solution of the given question.
Complete step by step answer:
The voltage supply is given as,\[V = 220V\]
Power consumed at minimum heating is given as, ${P_{\min }} = 360W$
Power consumed at maximum heating is given as, ${P_{\max }} = 840W$
We know that, $P = VI$…………….(i)
Also, we know from Ohm’s law, the resistance can be written as,
$R = \dfrac{V}{I}$…………………(ii)
Let us now find the value of current for minimum heating,
${P_{\min }} = VI$
$ \Rightarrow 360 = 220I$
$ \Rightarrow I = \dfrac{{360}}{{220}} = 1.63A$
Now, using equation (ii), we can find the resistance for minimum heating
$\Rightarrow R = \dfrac{{220}}{{1.63}} = 134.96\Omega $
Again current for maximum heating will be,
${P_{\max }} = VI$
$ \Rightarrow 840 = 220I$
$ \Rightarrow I = \dfrac{{840}}{{220}} = 3.81A$
Again by using equation (ii), we can find the resistance for maximum heating.
\[\Rightarrow R = \dfrac{{220}}{{3.81}} = 57.74\Omega \]
Therefore, the value of current for minimum power is $1.63A$ and the resistance is $134.96\Omega $.
The value of current for maximum power is $3.81A$ and the resistance is $57.74\Omega $.
Note: We define electric power as the rate per unit time at which there is a transfer of electrical energy by an electric circuit. Electric power can also be defined as the rate at which the work is done in an electric circuit. Its unit is ‘Watt’ or joule per second. Current is the rate of flow of charge or we can say the flow of charges per unit time.
Complete step by step answer:
The voltage supply is given as,\[V = 220V\]
Power consumed at minimum heating is given as, ${P_{\min }} = 360W$
Power consumed at maximum heating is given as, ${P_{\max }} = 840W$
We know that, $P = VI$…………….(i)
Also, we know from Ohm’s law, the resistance can be written as,
$R = \dfrac{V}{I}$…………………(ii)
Let us now find the value of current for minimum heating,
${P_{\min }} = VI$
$ \Rightarrow 360 = 220I$
$ \Rightarrow I = \dfrac{{360}}{{220}} = 1.63A$
Now, using equation (ii), we can find the resistance for minimum heating
$\Rightarrow R = \dfrac{{220}}{{1.63}} = 134.96\Omega $
Again current for maximum heating will be,
${P_{\max }} = VI$
$ \Rightarrow 840 = 220I$
$ \Rightarrow I = \dfrac{{840}}{{220}} = 3.81A$
Again by using equation (ii), we can find the resistance for maximum heating.
\[\Rightarrow R = \dfrac{{220}}{{3.81}} = 57.74\Omega \]
Therefore, the value of current for minimum power is $1.63A$ and the resistance is $134.96\Omega $.
The value of current for maximum power is $3.81A$ and the resistance is $57.74\Omega $.
Note: We define electric power as the rate per unit time at which there is a transfer of electrical energy by an electric circuit. Electric power can also be defined as the rate at which the work is done in an electric circuit. Its unit is ‘Watt’ or joule per second. Current is the rate of flow of charge or we can say the flow of charges per unit time.
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