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 nRn2RR/n2R/n

Solution

## The correct option is B n2RResistance 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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