# The equal resistance of 'n' resistors each of Resistance 'R' , Rs and Rp in series and parallel , prove that (Rs - Rp) = (n square - 1)R/n

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The equal resistance of 'n' resistors each of Resistance 'R' , Rs and Rp in series and parallel ,prove that

(Rs - Rp) = (n​2​- 1)R/n

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Solution:

We know that equivalent resistance in series connection is sum of all the individual resistances.

So, In series connection Rs = nR

Also The equivalent resistance in parallel connection is

1/Rp = 1/R+1/R+1/R+...................n times

=> 1/Rp = (1+1+1+1+........n times)/R

=> Rp = R/n

In parallel connection Rp = R/n

Now Rs - Rp = nR-R/n

=> Rs - Rp = (n2R-R)/n

=> Rs - Rp = (n​2​- 1)R/n  Proved

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Solution:

We know that equivalent resistance in series connection is sum of all the individual resistances.

So, In series connection Rs = nR  and

In parallel connection Rp = R/n

Now Rs - Rp = nR-R/n

=> Rs - Rp = (n2R-R)/n

=> Rs - Rp = (n​2​- 1)R/n  Proved