Two wires of equal diameters,lengths l_1, l_2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The resistance of a wire is given by $R = \frac{S l}{A}$,where $S$ is resistivity,$l$ is length,and $A$ is the cross-sectional area.Since the wire

Vedclass · Accepted Answer

$\frac{S_1 l_1+S_2 l_2}{l_1+l_2}$

Vedclass · Answer

(A) The resistance of a wire is given by $R = \frac{S l}{A}$,where $S$ is resistivity,$l$ is length,and $A$ is the cross-sectional area.Since the wires are joined in series,the total resistance $R$ is the sum of individual resistances: $R = R_1 + R_2$.Substituting the formula for resistance: $\frac{S(l_1 + l_2)}{A} = \frac{S_1 l_1}{A} + \frac{S_2 l_2}{A}$.Since the diameters are equal,the cross-sectional area $A$ is the same for both wires.Canceling $A$ from both sides,we get: $S(l_1 + l_2) = S_1 l_1 + S_2 l_2$.Therefore,the equivalent resistivity $S$ is: $S = \frac{S_1 l_1 + S_2 l_2}{l_1 + l_2}$.

Two wires of equal diameters,lengths $l_1, l_2$ and having resistivities $S_1, S_2$ respectively are joined in series. The equivalent resistivity of the combination is

Similar Questions

At what temperature $(^{\circ} C)$ will the resistance of a copper wire become $3$ times its resistance at $0^{\circ} C$? (Given: Temperature coefficient of resistance $\alpha = 4 \times 10^{-3} {}^{\circ} C^{-1}$)

If the length of a wire is doubled, its specific resistance (resistivity) will be.......

If the length of a wire is increased by $10\%$ by stretching it,the percentage increase in its resistance is ..............$\%$

$A$ strip of copper and another of germanium are cooled from room temperature to $80\, K$. The resistance of

The length of the wire is doubled. Its conductance will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module