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Question

Resistance of n resistors each of r Ω, when connected in parallel give an equivalent resistance of R Ω. If these resistance's were connected in series, the combination would have a resistance in Ω, equal to 
  1. nR
  2. n2R
  3. R/n2
  4. R/n


Solution

The correct option is B n2R
Resistance in parallel,
1R=1R1+1R2++1Rn
If R1=R2==Rn=r
1R=1r+1r++n
1R=nr
r=nR(i) 
Now resistance in series,
R=R1+R2++Rn
R=r+r++n
R=nr
From Eq.(i) put r value in above equation
R=n(nR)=n2R

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