Question:
Let there be n resistors R1……..Rn with Rmax = max(R1……Rn) and Rmin = min(R1…….Rn). Show that when they are connected in parallel, the resultant resistance Rp < Rmin and when they are connected in series, the resultant resistance Rs > Rmax. Interpret the result physically.
Solution:
Following is the parallel grouping of the resistance:
Rp = 1/R1 + 1/R2 + …. + 1/Rn
Rmin/Rp = Rmin/R1 + Rmin/R2 + …..+Rmin/Rn
The RHS is equal to 1
Therefore, Rp < Rmin
Following is the series grouping of the resistance:
Rs = R1 + R2 + …… + Rn
Rs = R1 + ….. +Rmax+…..+Rn
Rs = Rmax(R1+……..Rn)
Therefore, series combination has an equivalent resistance of resistors which is greater than the maximum resistance in parallel combination.
Physical interpretation is: