An electric fuse is rated 5 A. An electric iron rated 1.5 kW operates on a 220 V supply. Will this fuse prote…

Question

An electric fuse is rated 5 A. An electric iron rated 1.5 kW operates on a 220 V supply. Will this fuse protect the electric iron safely? Justify your answer with a calculation.

Accepted Answer

## Model Answer **Calculation:** Power of electric iron, P = 1.5 kW = 1500 W Voltage, V = 220 V Current drawn by the iron: $$I = \frac{P}{V} = \frac{1500}{220} \approx 6.82 	ext{ A}$$ **Conclusion:** No, the 5 A fuse will **not** protect the electric iron safely. Since the iron draws approximately 6.82 A, which is greater than the fuse rating of 5 A, the fuse will blow (melt) and break the circuit every time the iron is used. A fuse of higher rating (e.g., 7 A or 10 A) should be used instead. *Source: Chapter 11, Section 11.7.1* --- ## Explanation - The key formula is $I = P/V$. Convert kW to W first (1.5 kW = 1500 W). - Since the calculated current (6.82 A) **exceeds** the fuse rating (5 A), the fuse will blow during normal operation — so it does **not** protect the iron safely; it simply keeps tripping. - Examiners expect: the formula, correct substitution, correct answer (~6.82 A), and a clear justified conclusion. All three steps carry marks.

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