​A house is supplied electricity through a 15 A fuse. The number of 100 W lamps that can be used simultaneously along with a 2KW AC is: (The AC and lamps both are rated for 220 V Supply)​

Correct option is D

The correct answer is (D) 13.

• Electric Power:
• The rate at which energy is converted into an electrical circuit or work is completed is known as electric power. To put it another way, it is a measure of how much energy is used over a time period.
• The formula for electric power is given by P=VI where V is the potential difference in the circuit and P is the power.
• I is the current of electricity.
• Power can also be written as P=I²R and P= $$\frac{V^2}R$$​
• Ohm's law is used to get the above two expressions, where voltage, current, and resistance are related in the following way: V=IR

1.Calculate the current drawn by the AC:

• Power of AC (P) = 2 kW = 2000 W
• Voltage (V) = 220 V
• Current (I) =$$\frac{ P} V = \frac{2000 W}{220 V}$$​ ≈ 9.09 A

2. Calculate the maximum current available for lamps:

• Fuse rating = 15 A
• Current for AC = 9.09 A
• Maximum current for lamps = 15 A - 9.09 A ≈ 5.91 A

3. Calculate the current drawn by each lamp:

• Power of each lamp (P) = 100 W
• Voltage (V) = 220 V
• Current per lamp (I) = $$\frac{P } V = \frac{100 W }{ 220 V}$$​ ≈ 0.45 A

4. Calculate the maximum number of lamps:

• Maximum current for lamps = 5.91 A
• Current per lamp = 0.45 A
• Maximum number of lamps = $$\frac{5.91 A} {0.45 A}$$​ ≈ 13.13

Since we can't have a fraction of a lamp, the maximum number of 100 W lamps that can be used simultaneously with the 2 kW AC is 13.