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Grade: Upto college level
        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? 
7 years ago

Answers : (7)

Chetan Mandayam Nayakar
312 Points
							

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

parallel:rho(parallel)=(rho1)(rho2)/(rho1 + rho2)

7 years ago
Ritvik Agarwal
8 Points
							

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)

7 years ago
Chetan Mandayam Nayakar
312 Points
							

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

7 years ago
anurag bhattacharjee
39 Points
							

yes sir ,i want to know how to approach and solve this question exactly instead of just knowing the answers.

7 years ago
xyz xz
37 Points
							

R1= ρ1(l/A)

R2= ρ2(l/A)

R= ρ(2l)/A

and Rs = R1+R2

» ρs = (ρ12)/2

and Rp = (R1.R2)/(R1+R2)

» ρp = (1/2)(ρ1ρ2)/(ρ1+ρ2)

7 years ago
mahima
13 Points
							
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
one year ago
Gitanjali Rout
184 Points
							
 

R1= ρ1(l/A)

R2= ρ2(l/A)

R= ρ(2l)/A

and Rs = R1+R2

» ρs = (ρ12)/2

and R= (R1.R2)/(R1+R2)

» ρp = (1/2)(ρ1ρ2)/(ρ1+ρ2)

one year ago
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