Step 1: Understanding the Concept:
The fuse is a safety device that prevents excessive current from flowing through an appliance.
The rating of a fuse is determined by the maximum current the appliance draws during normal operation.
Step 2: Key Formula or Approach:
Electrical Power (\( P \)) is the product of Voltage (\( V \)) and Current (\( I \)).
\[ I = \frac{P}{V} \]
Step 3: Detailed Explanation:
Given:
Power (\( P \)) = 1100 W
Voltage (\( V \)) = 220 V
Substituting the values into the formula:
\[ I = \frac{1100}{220} \]
\[ I = 5 \text{ A} \]
The current drawn by the electric iron is 5 A. Therefore, a fuse with a minimum rating of 5 A is required to allow the appliance to function normally.
Step 4: Final Answer:
The minimum rating of the fuse required is 5 A.
(b)
Step 1: Understanding the Concept:
Electrical energy is the total power consumed over a specific period. The commercial unit for electrical energy is the kilowatt-hour (kWh).
Step 2: Key Formula or Approach:
Energy (\( E \)) in kWh = Power (\( P \)) in kW \(\times\) Time (\( t \)) in hours.
Step 3: Detailed Explanation:
First, convert the power from Watts to Kilowatts:
\[ P = \frac{1100}{1000} \text{ kW} = 1.1 \text{ kW} \]
Given time \( t = 5 \text{ hours} \).
Calculate Energy:
\[ E = 1.1 \text{ kW} \times 5 \text{ h} \]
\[ E = 5.5 \text{ kWh} \]
Step 4: Final Answer:
The energy consumed by the electric iron is 5.5 kWh.
(c)
Step 1: Understanding the Concept:
One "unit" of electricity as billed by utility companies is exactly 1 kWh. The total cost is the product of units consumed and the rate per unit.
Step 2: Detailed Explanation:
From the previous calculation, the energy consumed is 5.5 kWh.
Units consumed = 5.5 units.
Rate per unit = \₹ 10.
\[ \text{Total Cost} = \text{Units} \times \text{Rate} \]
\[ \text{Total Cost} = 5.5 \times 10 = 55 \]
Step 3: Final Answer:
The total cost of the energy consumed is ₹ 55.