Electricity — Question 12

Given,

V = 220 V

P = 10 W

I_Max = 5 A

The resistance of the each bulb:

R = $\dfrac{\text{V}^2}{\text{P}} = \dfrac{220^2}{10} = 4840 Ω$

Assuming n bulbs are connected in parallel

$\dfrac{1}{\text{R}_{eq}} = \dfrac{1}{\text{R}} + \dfrac{1}{\text{R}} + ....n \\[1em] \dfrac{1}{\text{R}_{eq}} = \dfrac{1}{4840} + \dfrac{1}{4840} + ....n \\[1em] \dfrac{1}{\text{R}_{eq}} = \dfrac{\text{n}}{4840} \Omega \\[1em] \Rightarrow \text{R}_{eq} = \dfrac{4840}{\text{n}} \Omega$

Applying Ohm's Law:

V = IR

220 = 5 x $\dfrac{4840}{\text{n}}$

n = $\dfrac{4840 \times 5}{220}$ = 110 lamps

Hence, 110 lamps can be connected in parallel.