Q.
Two wires of equal diameters, of resistivities ρ1 and ρ2 and length l1 and l2 , respectively are joined in series. The equivalent resistivity of the combination is
4272
186
NTA AbhyasNTA Abhyas 2020Current Electricity
Report Error
Solution:
Wires having frame diameter, hence area of cross-section will be equal,
i.e. A1=A2=A R1=ρ1.Al1
and R2=ρ2.Al2
Since, both resistance are joined in series.
Hence, Req=R1+R2 =ρ1⋅Al1+ρ2⋅Al2 Req=A1(ρ1l1+ρ2l2) ρeq⋅A(l1+l2)=A1(ρ1l1+ρ2l2) ⇒ρeq=l1+l2ρ1l1+ρ2l2