An electric iron rated 1100 W, 220 V is operated for 5 hours.

Given,

• Power rating of the iron ($\text P$) = 1100 W
• Voltage rating of the iron ($\text V$) = 220 V
• Operating time = 5 hours

(a) Power consumed by an appliance is given by,

$$\begin{aligned}\text P = \text {VI} \\[1em] \Rightarrow \text I = \frac{\text P}{\text V} \\[1em] = \frac{1100}{220} \\[1em] = 5\ \text A\end{aligned}$$

Here, $\text I$ is the operating current flowing through the appliance.

Since the fuse rating should be slightly higher than the operating current, so the minimum suitable fuse rating is 6 A.

Hence, 6 A is the minimum rating of the fuse required.

(b) Energy consumed by the iron = Power consumed x Operating time

= 1100 x 5

= 5500 Wh

= 5.5 x 10³ Wh

= 5.5 kWh

Hence, the energy consumed by the iron is 5.5 kWh.

(c) Given,

• Electricity rate = ₹ 10 per unit

As, 1 unit = 1 kWh

Then,

The cost of energy consumed by the iron = Energy consumed x Electricity rate

= 5.5 x 10

= ₹ 55

Hence, the cost of the energy consumed is ₹ 55.