Solution:P = V × I (Where P = power, V = Voltage, I = Current) V
= 220 V, I = 15 A, Power of AC = 2Kw = 2000 watt
Now, total power P = VI ⇒ P = 220 × 15 = 3300 W
Let n be no of bulbs can be used in the circuit, then
𝑃𝑜𝑤𝑒𝑟ₜₒₜₐₗ = 𝑃₁+ 𝑃₂ ⇒ P = n × 𝑝₁ + 1 𝑝₂
3300= n × 100 + 200
100n = 3300 - 2000 ⇒ 100n = 1300
⇒ n = 1300/100 = 13