Question

A 9 ohm resistance wire is doubled on itself .Calculate the new resistance of the wire.​

Answer 1

The new resistance of the wire is 1.5 ohm.

Step-by-step explanation:

Given : A 6 ohm resistance wire is doubled on itself.

To find : Calculate the new resistance of the wire. ?

Solution :

The resistance is given by,

Here, Resistance=6 ohm

In second case,

Resistance is  

As the wire is doubled,

So  and  

Therefore, the new resistance of the wire is 1.5 ohm.

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A 4 ohm resistance wire is doubled on itself calculate the new resistance of the wire

Answer 2

Answer:

Explanation: very easy

first of all R1 = 9Ω

                we know that  R1 = p x length of wire/ cross section area

                                           R1 = p l/A ___________  [1]

             

      here , wire is doubled on itself means that a single wire is folded.

   As we know if it is folded then of course it will become thick and

   short.

   so length { l } will be halved as it is double folded so l = l/2

    and wire will become thick so cross section area will increase , it is

    double folded so Area { A} = 2A

    let new resistance be R2

     Therefore,

   R2 = p x l/A

    R2 =  p x l/2/2A                                  { l = l/2 and A = 2A }

   

   R2 =  p x l/4A

   R2  = 1/4  { p x l/A}

   R2 =  1/4 { R1 }                              _____ from 1 that {R1 = p x l/A}

 

   R2  = 1/4 x 9Ω             ------{ R1 = 9Ω - given }

   

  R2 = 2.25Ω

 

 so new resistance is 2.25Ω

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