Question

N conducting wires of same dimensions, but having resistivities 1, 2, 3,…n are connected in series. the equivalent resistivity of the combination is

Answer 1

Final Answer : Equivalent resistivity of Combination.
$\frac{n(n + 1)}{2}$

Steps and Understanding:

1) Resistance of a wire is given by :
pl/A
where p = resistivity of wire
l = length of wire.

2) Since, Dimension of all wires are same so
l, A are also same.
p(1) = 1
p(2) = 2
p(3)= 3, ... p(n) = n.

Here, by a equivalent resistivity , I mean that resistivity of a Resistor having dimension same as other resistors and same potential difference passes a same current through R(eff).


All resistors are in series , So
R(eq) = p(1) l/A. + p(2) l/A + p(3)l/A ...... p(n)l/A.
=> p(eq)l/A = l/A ( 1 + 2 +3 + .... n)
=> p(eq) = 1+2+3+...n
=> p(eq) = n(n+1) /2 .

Hence, Resistivity of combination is n(n+1)/2.

Answer 2

Explanation:

hope it helps u