Read the following information and answer the questions that follow:
A household has the following electrical appliances all connected in parallel to a 220 V mains supply: a 1100 W electric iron, a 100 W television, and a 60 W table fan. The household uses a single fuse in the main line to protect the circuit.
(a) Calculate the total current drawn from the mains when all three appliances are operating simultaneously. [1 mark]
(b) The family has fuses of ratings 5 A and 10 A available. Which fuse should be used in the main line, and why? [1 mark]
(c) Explain briefly how a fuse wire protects the appliances in a circuit when excessive current flows. [1 mark]
(d) If the electric iron is operated daily for 2 hours and the cost of electricity is ₹3.00 per kWh, what is the cost of running the iron for 30 days? [1 mark]
Generated by claude-sonnet-4-6 · 2026-06-26 01:10 · grounding rag
Model Answer
(a) Total power = 1100 + 100 + 60 = 1260 W
Total current, $I = \dfrac{P}{V} = \dfrac{1260}{220} \approx 5.73 \text{ A}$
(b) The 10 A fuse should be used. The total current drawn is ~5.73 A, which exceeds the 5 A fuse rating, so a 5 A fuse would blow immediately. A 10 A fuse allows normal operation while still protecting against dangerously excessive current.
(c) A fuse wire is made of a metal/alloy with a low melting point, connected in series. When current exceeds the rated value, the fuse wire heats up, melts, and breaks the circuit — stopping current flow and protecting the appliances.
(d) Energy consumed = 1100 W × 2 h × 30 days = 66,000 Wh = 66 kWh
Cost = 66 × ₹3.00 = ₹198.00
Source: Chapter 11 – Electricity, Section 11.7.1 (Fuse) and Section 11.8 (Electric Power)
---
Explanation
- (a) Use $I = P/V$ for total power since all appliances are in parallel across the same 220 V.
- (b) Key rule from the textbook: fuse rating must be above normal current but as low as safely possible. 5 A would blow under normal use; 10 A is correct. The textbook example of a 1000 W iron drawing 4.54 A uses a 5 A fuse — here total current is higher, so 10 A is needed.
- (c) Mention: series connection, low melting point, melts on excess current, breaks circuit. These are the examiner's expected points.
- (d) Follow the textbook's Example 11.13 method exactly: Energy (kWh) = Power (kW) × time (h) × days, then multiply by rate.