Two wires of same radius having lengths l1 and l2 and resistivities ρ1 and ρ2 are connected in series. The equivalent resistivity will be

Question

Two wires of same radius having lengths l1 and l2 and resistivities ρ1 and ρ2 are connected in series. The equivalent resistivity will be (A) (ρ1 l2+ρ2 l1/ρ1+ρ2) (B) (ρ1 l1+ρ2 l2/l1+l2) (C) (ρ1 l1+ρ2 l2/l1-l2) (D) (ρ1 l2+ρ2 l1/l1+l2)

Tardigrade Admin · Accepted Answer

In series equivalent resistance R=R1+R2 R1=(ρ1 l1/A), R2=(ρ2 l2/A) Hence, R=(ρ(l1+l2)/A) ∴ (ρ(l1+l2)/A)=(ρ1 l1/A)+(ρ2 l2/A) ⇒ ρ=(ρ1 l1+ρ2 l2/l1+l2)

Q. Two wires of same radius having lengths $l_{1}$ and $l_{2}$ and resistivities $ρ_{1}$ and $ρ_{2}$ are connected in series. The equivalent resistivity will be

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

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

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

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

Q. Two wires of same radius having lengths l1​ and l2​ and resistivities ρ1​ and ρ2​ are connected in series. The equivalent resistivity will be

ρ1​+ρ2​ρ1​l2​+ρ2​l1​​

l1​+l2​ρ1​l1​+ρ2​l2​​

l1​−l2​ρ1​l1​+ρ2​l2​​

l1​+l2​ρ1​l2​+ρ2​l1​​

In series equivalent resistance R=R1​+R2​ R1​=Aρ1​l1​​,R2​=Aρ2​l2​​ Hence, R=Aρ(l1​+l2​)​ ∴Aρ(l1​+l2​)​=Aρ1​l1​​+Aρ2​l2​​ ⇒ρ=l1​+l2​ρ1​l1​+ρ2​l2​​

Q. Two wires of same radius having lengths $l_{1}$ and $l_{2}$ and resistivities $ρ_{1}$ and $ρ_{2}$ are connected in series. The equivalent resistivity will be

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

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

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

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