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(a) An electric iron is rated at 230 V, 750 W. Calculate (i) the maximum current, and (ii) the number of units of electricity it would use in 30 minutes.
(b) Which of the following fuse ratings would be suitable for this electric iron?
1 A, 3 A, 5 A, 13 A
(a) (i) Given,
Potential difference, V = 230 V
Power, P = 750 W
Current, I = ?
Let the maximum current be I.
We know that,
$P=V\times I$
Then, current $I=\frac{P}{V}$
Substituting the given values in the formula, we get-
$I=\frac{750}{230}$
$I=3.26A$
Thus, the value of the maximum current is 3.26 A.
(ii) Given:
Power, P = 750 W = 0.750 kW [converted watt into kilowatt by dividing from 1000]
Time, t = 30 minutes = $\frac{30}{60}hours=\frac{1}{2}h$ [converted minutes into hour by dividing from 60]
Energy (no. of units of electricity), E = ?
We know that Electric energy consumed, is given as:
$E=P\times t$
Substituting the given values in the formula, we get-
$E=0.750\times \frac{1}{2}$
$E=0.750\times \frac{1}{2}$
$E=0.375kWh\ or\ 0.375units$
Thus, 0.375 units of electricity it would use in 30 minutes.
(b) We've found that the normal current drawn by the electric iron is 3.26 A, so a fuse rating of 5 A would be suitable for this because the fuse used by an appliance should be slightly more than the normal current drawn through it.
Updated on: 10-Oct-2022
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