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two wires of equal length but different cross sectional area A 1 and A 2 and specific resistance ρ 1 and ρ 2 are joined in parallel .the equivalent specific resistance is given as___(ans:ρ 1 ρ 2 (A 1 +A 2 )/[ρ 1 A 2 +ρ 2 A 1 ])

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4 years ago

```							We have length of resistances R1 and R2 = l.Area of cross section A1 and A2 resp of R1 and R2Specific resistances ρ1 and ρ2 resp of R1 and R2We have Req = (R1.R2)/(R1+R2)Where R1= ρ1.l/A1            and      R2 = ρ2.l/A2Putting both values in equationReq =( ρ1.l/A1).( ρ2.l/A2)/( ρ2.l/A2+ ρ1.l/A1)Req =l(ρ1.ρ2)/( ρ1.A2+ ρ2.A1)Then we know Aeq = A1+A2So Req = ρeq.l/Aeq                    (length is constant overall)So ρeq = (A1+A2). ρ1.ρ2/( ρ1.A2+ ρ2.A1)Hence proved.
```
4 years ago
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