An electric iron consumes energy at a rate of $840\, W$ when heating is at the maximum rate and $360\, W$ when the heating is-at the minimum. The voltage is $220\, V$. What are the current and the resistance in each case ?
An electric iron consumes energy at a rate of $840\, W$ when heating is at the maximum rate and $360\, W$ when the heating is-at the minimum. The voltage is $220\, V$. What are the current and the resistance in each case ?
We know that the power input is $P=V I$
Thus, the current $I=P / V$
$(a)$ When heating is at the maximum rate,
$I=840 / 220=3.82 A$
and the resistance of the electric iron is
$R = V / I =220 / 3.82=57.60 \Omega$
$(b)$ When heating is at the minimum rate,
$I=360 / 220=1.64 A$
and the resistance of the electric iron is
$R = V / I =220 / 1.64=134.15 \Omega$
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