2 wires of same dimension but different resistivity(rho 1 & rho 2) are connected in series and after that in parallel,then the equivalent resistivity of the conductor in both the case will be?

							
series: rho(series)= (rhoi+rho2)/2
parallel:rho(parallel)=(rho1)(rho2)/(rho1 + rho2)
						

							
R1=(rho1)L/A
R2=(rho2)L/A
In series; Req=(rho1+rho2)L/A =>Resistivity=(rho1+rho2)
In parallel; Req=((rho1)(rho2)/(rho1+rho2))L/A =>Resistivity=Req(parallel) X (A/L)
						

							
my previous solution was wrong. the correct answer is:
series: (rho1 + rho2)/2
parallel: (rho1*rho2)/(2*(rho1+rho2))
please inform me in case you want more explanation
						

							
R1= ρ1(l/A)
R2= ρ2(l/A)
R= ρ(2l)/A
and Rs = R1+R2
» ρs = (ρ1+ρ2)/2
and Rp = (R1.R2)/(R1+R2)
» ρp = (1/2)(ρ1ρ2)/(ρ1+ρ2)
						

							
First we can start with calculating the equivalent resisitance of the combination.  {R = rho*l/a}
rho*2*l/a = (rho 1+ rho 2)l/a
in this case since the dimensions of both the wires are same, the total length is considered for net resistance
So l/a gets cancelled on both sides and we have..
rho(total) *2 = rho 1 + rho 2
this implies that,
rho (total) = (rho 1 + rho 2)/2
						

							
 
R1= ρ1(l/A)
R2= ρ2(l/A)
R= ρ(2l)/A
and Rs = R1+R2
» ρs = (ρ1+ρ2)/2
and Rp = (R1.R2)/(R1+R2)
» ρp = (1/2)(ρ1ρ2)/(ρ1+ρ2)