Question

a circuit has aline of 5ampere .how many lamps of rating 40W and 220V can simultaniously run on this line safely?

Answer 1

$\blue{These\: lamps\: are\: arranged\: in\: series-}$
$\red{\boxed{\huge{nP=VI}}}$

Here,
n=no. of Bulbs
P=power used
V=voltage
I=current in circuit.

$n \times 40 = 220 \times 5 \\ n = \frac{220 \times 5}{4} \\ n = 27.5$

$\orange{\:number\:of\:object\: can't\:be\:\\in\:decimal}$

Hence answer will be-

$\boxed{\huge{Answer=27}}$

Answer 2

$Here \: is \: the \: answer \: of \: your \: question$

This question can be solved using Two Methods.

`$\textbf{First Method}$`

→ According to question we have given,

P (power) = 40 W

Volt (V) = 220 V

R1 (resistance) = ?

R1 = V²/P

R1 = [(220 × 220) ÷ 40]

R1 = 1210 ohm

Now,

max. volt (V) = 220 V

and max. current (I) = 5 A

So, max. R1, when (n) number of resistance are connected in parallel.

R2 = V/I

R2 = 220/5 = 44 ohm

Now,

R2 = r1/n

44 = 1210/n

n = 27.5

On rounding off we get,

`$\textbf{number of lamps (n) = 27}$`

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`$\textbf{Second Method}$`

→ current (I) = ?

power (P) = 40 W

volt (V) = 220 V

So,

I = P/V

I = 40/220 = 0.18 A of one lamp

Max. I = 5 A

So,

Number of lamp = [Max. current (I) ÷ Current pass through one lamp]

= (5 ÷ 0.18)

= 27

So,

`$\textbf{Number of lamps = 27}$`

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