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 $360W$ but at ‘maximum heating’ it takes a power of $840W$. Calculate the current and resistance in each case.
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Given: $V = 220V, P_{min} = 360W, P_{max} = 840W$
For minimum heating case:
We know that
$P_{min} = VI$
$360 = 220XI$
$I = 1.63 amp$
$\text{R}=\frac{\text{V}}{\text{I}}$
$\text{R}=\frac{220}{1.63}$
$R = 134.96\ ohms$
For maximum heating case:
We know that
$P_{max} = VI$
$840 = 220XI$
$I = 3.81 amp$
$\text{R}=\frac{\text{V}}{\text{I}}$
$\text{R}=\frac{220}{3.81}$
$R = 57.74\ ohms$
For minimum heating case:
We know that
$P_{min} = VI$
$360 = 220XI$
$I = 1.63 amp$
$\text{R}=\frac{\text{V}}{\text{I}}$
$\text{R}=\frac{220}{1.63}$
$R = 134.96\ ohms$
For maximum heating case:
We know that
$P_{max} = VI$
$840 = 220XI$
$I = 3.81 amp$
$\text{R}=\frac{\text{V}}{\text{I}}$
$\text{R}=\frac{220}{3.81}$
$R = 57.74\ ohms$
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