Two wires of equal diameters, of resistivities ρ 1 and ρ 2 and length l1 and l2 , respectively are joined in series. The equivalent resistivity of the combination is

Question

Two wires of equal diameters, of resistivities ρ 1 and ρ 2 and length l1 and l2 , respectively are joined in series. The equivalent resistivity of the combination is (A) (ρ 1 l1 + ρ 2 l2/l1 + l2) (B) (ρ 1 l2 + ρ 2 l1/l1 - l2) (C) (ρ 1 l2 + ρ 2 l1/l1 + l2) (D) (ρ 1 l1 + ρ 2 l2/l1 - l2)

Tardigrade Admin · Accepted Answer

Wires having frame diameter, hence area of cross-section will be equal, i.e. A1=A2=A R1=ρ 1.(l1/A) and R2=ρ 2.(l2/A) Since, both resistance are joined in series. Hence, Re q=R1+R2 =ρ1 ⋅ (l1/A)+ρ2 ⋅ (l2/A) Re q=(1/A)(ρ1 l1+ρ2 l2) ρe q ⋅ ((l1+l2)/A)=(1/A)(ρ1 l1+ρ2 l2) ⇒ ρe q=(ρ1 l1+ρ2 l2/l1+l2)

Q. Two wires of equal diameters, of resistivities $ρ_{1}$ and $ρ_{2}$ and length $l_{1}$ and $l_{2}$ , respectively are joined in series. The equivalent resistivity of the combination is

$\frac{ρ _{1} l _{1} + ρ _{2} l _{2}}{l _{1} + l _{2}}$

$\frac{ρ _{1} l _{2} + ρ _{2} l _{1}}{l _{1} - l _{2}}$

$\frac{ρ _{1} l _{2} + ρ _{2} l _{1}}{l _{1} + l _{2}}$

$\frac{ρ _{1} l _{1} + ρ _{2} l _{2}}{l _{1} - l _{2}}$

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

l1​+l2​ρ1​l1​+ρ2​l2​​

l1​−l2​ρ1​l2​+ρ2​l1​​

l1​+l2​ρ1​l2​+ρ2​l1​​

l1​−l2​ρ1​l1​+ρ2​l2​​

Q. Two wires of equal diameters, of resistivities $ρ_{1}$ and $ρ_{2}$ and length $l_{1}$ and $l_{2}$ , respectively are joined in series. The equivalent resistivity of the combination is

$\frac{ρ _{1} l _{1} + ρ _{2} l _{2}}{l _{1} + l _{2}}$

$\frac{ρ _{1} l _{2} + ρ _{2} l _{1}}{l _{1} - l _{2}}$

$\frac{ρ _{1} l _{2} + ρ _{2} l _{1}}{l _{1} + l _{2}}$

$\frac{ρ _{1} l _{1} + ρ _{2} l _{2}}{l _{1} - l _{2}}$