Question

An electrical bulb is rated 40 w, 220 v. how many bulbs can be connected in parallel with each other across the two wires of 220 v line if the maximum allowable current is 6 a?

Answer 1

first find the current of 1 bulb
I=p/v
I=40/220=2/11
The total current =6a divided by the current of 1 bulb.
total no. of bulbs=6/2/11=33
total no. of bulbs=33

Answer 2

Answer:

33 (approx.)

Explanation:

Given,

P=40 W

V=220 V

I=6 A

We know that,

$\mathsf{ R = \dfrac{V^2}{P}}$

$\mathsf{= \dfrac{(220)^2}{40}}$

$\mathsf{= \dfrac{\cancel{48400_{1210}}}{\cancel{40_{1}}}}$

$\mathsf{= 1210Ω}$

Now,

$\mathsf{V=IR_{eq}}$

$\mathsf{⇒ R_{eq}= \dfrac{V}{I} }$

$\mathsf{⇒ \dfrac{220}{6}}$

$\mathsf{= 36.66Ω}$

Suppose, there are x number of bulb in parallel

$\mathsf{36.66 = \dfrac{1210}{x}}$

$\mathsf{x = \dfrac{1210}{36.66}}$

$\mathsf{x = 36.066}$

$\mathsf{x = 33 (approx.)}$