Wednesday 15 May 2013

An electric lamp of 100 Ω a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω



Question 2:
An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Solution:
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 
 

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,
 
 






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

 

21 comments:

  1. Wow really helpful

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  2. Wish I had this website during my 10th.

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  3. Hi,he mentioned electric iron also connected.then how can u take only R1,R2,R3 and calculate total resistance of a ckt

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  4. Really this helped me a lot..
    Thanks a lot

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  5. It helps me to understand. I am sure that it helps to everyone.

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  6. Why we reciprocal the 16 by 500 and again reciprocal it to 500 bye 16

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    Replies
    1. Because it is connected in parallel. That is y

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    2. Because it is connected in parallel. That is y

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  7. 7.04 is the total current ain't It?? So all three appliances are taking it collectively???

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  8. Thoda aur zyada simple tarika batao

    ReplyDelete