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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?

```
8 years ago

```							series: rho(series)= (rhoi+rho2)/2
parallel:rho(parallel)=(rho1)(rho2)/(rho1 + rho2)
```
8 years ago
```							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)
```
8 years ago
```							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
```
8 years ago
```							yes sir ,i want to know how to approach and solve this question exactly instead of just knowing the answers.
```
8 years ago
```							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)
```
7 years ago
```							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/ain this case since the dimensions of both the wires are same, the total length is considered for net resistanceSo l/a gets cancelled on both sides and we have..rho(total) *2 = rho 1 + rho 2this implies that,rho (total) = (rho 1 + rho 2)/2
```
2 years ago
```							 R1= ρ1(l/A)R2= ρ2(l/A)R= ρ(2l)/Aand Rs = R1+R2» ρs = (ρ1+ρ2)/2and Rp = (R1.R2)/(R1+R2)» ρp = (1/2)(ρ1ρ2)/(ρ1+ρ2)
```
2 years ago 605 Points
```							Dear student,Please find the solution to your problem below. R1= ρ1(l/A)R2= ρ2(l/A)R= ρ(2l)/A In series, Rs = R1+R2→ ρs = (ρ1+ρ2)/2 In parallel, Rp = (R1.R2)/(R1+R2)→ ρp = (1/2)(ρ1ρ2)/(ρ1+ρ2) Hope it helps.Thanks and regards,Kushagra
```
3 months ago
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