Question

an electric lamp and a toaster and a water filter has 100 ohm and 50 ohm and 500 ohm respectively are connected in a parralel to a 220 volt source. what is the resistence of an electric iron connected to the same source that take as much as current as all three appliances and what is the currwnt throigh it?

Answer 1

Given that Resistance of an Electric lamp R1 = 100 ohms.

Given that Resistance of a toaster R2 = 50 ohms.

Given that Resistance of a water filter = 500 ohms.

Given that Potential difference of the source = 220V.

Given that they are connected in parallel.

Let R be the equivalent resistance of the circuit.

1/R = 1/R1 + 1/R2 + 1/R3

      = 1/100 + 1/50 + 1/500


LCM of 100,50,500 = 500.


1/R = 5 + 10 + 1/500

      = 16/500.


R = 500/16   ----------------------- (1)



We know that According to Ohm's law,

I = V/R

  = 220/500/16

  = 220 * 16/500

  = 3520/500

  = 7.0A.


Now,

Resistance of an Electric Iron = V/I

                                                  = 220/7.04

                                                  = 31.25 ohms.



Therefore the resistance of an electric iron = 31.25ohms and current = 7.04 A.


Hope this helps!  -------------- Gud Luck

Answer 2

HELLO DEAR,



Resistance of electric lamp (R1) = 100 Ω

Resistance of toaster (R2) = 50 Ω

Resistance of water filter (R3) = 500 Ω

Potential difference of the source, V = 220 V

Since all the resistance are in parallel then the equivalent resistance R of the circuit will be

$\frac{1}{r} = \frac{1}{ r_{1} } + \frac{1}{ r_{2} } + \frac{1}{ r_{3} } \\ = > \frac{1}{r} = \frac{1}{100} + \frac{1}{50} + \frac{1}{500} \\ = > \frac{5+ 10 + 1}{500} = \frac{16}{500} \\ = > r = \frac{500}{16} \\ = >$
According to ohms law
$i = \frac{v}{r}$
where current following through the circuit =I

$i = \frac{220}{ \frac{500}{16} } \\ = > \frac{220 \times 16}{500} = 7.04 \: a$
All the three given appliances are drawing 7.04 A of current.

Therefore, current drawn by an electric iron connected to the same source of potential 220 V = 7.04 A

Let   Rʹ be the resistance of the electric iron. According to Ohm’s law,


V=IR'

R' =V/I.

=>220/7.04 =31.25ohm

Therefore, the resistance of the electric iron is 31.25 Ω and the current flowing through it is 7.04 A.

I HOPE ITS HELP YOU DEAR
THANKS