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Question
An electric iron is connected to the mains power supply of 220 V. 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.
Solution
In the case of minimum heating:
P = 360 W
V = 220 V
where the symbols used have their usual meanings.
We know that P = VI
`I=p/V`
`I=360/220=1.64 A`
Resistance,`R V/I`
`R=220/1.64=134.15`Ω
In the case of maximum heating:
P = 840 W
V = 220 V
where the symbols used have their usual meanings.
We know that P = VI
`I=P/V`
`I=840/220=3.82A`
further,`R=V/I`
`R=220/3.82=57.60`Ω
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