Question
Two wires of equal diameters of resistivity's $\rho_{1}$ and $\rho_{2}$ are joined in series. The equivalent resistivity of the combination is....(A) $\left\{\left(\rho_{1} \ell_{1}+\rho_{2} \ell_{2}\right) /\left(\ell_{1}+\ell_{2}\right)\right\}$(B) $\left\{\left(\rho_{1} \ell_{2}+\rho_{2} \ell_{1}\right) /\left(\ell_{1}-\ell_{2}\right)\right\}$(C) $\left\{\left(\rho_{1} \ell_{2}+\rho_{2} \ell_{1}\right) /\left(\ell_{1}+\ell_{2}\right)\right\}$(D) $\left\{\left(\rho_{1} \ell_{1}+\rho_{2} \ell_{2}\right) /\left(\ell_{1}-\ell_{2}\right)\right\}$
Step 1
Step 1: The resistance of a wire is given by the formula $R=\rho \frac{l}{A}$, where $\rho$ is the resistivity, $l$ is the length, and $A$ is the cross-sectional area of the wire. Show more…
Show all steps
Your feedback will help us improve your experience
Prem Bijarniya and 70 other Physics 102 Electricity and Magnetism educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Two wires of equal lengths, equal diameters and having resistivities $\rho_{1}$ and $\rho_{2}$ are connected in series The equivalent resistivity of the combination is.... (A) $\left(\rho_{1}+\rho_{2}\right)$ (B) $(1 / 2)\left(\rho_{1}+\rho_{2}\right)$ (C) $\left\{\left(\rho_{1} \rho_{2}\right) /\left(\rho_{1}+\rho_{2}\right)\right\}$ (D) $\left.\sqrt{(} \rho_{1} \rho_{2}\right)$
Two wires of the same dimensions but different resistivity p1 and p2 are connected in series. The equivalent resistivity of the combination is: A. p1 + p2/2 B. β(p1p2) C. p1 + p2 D. (p1 + p2)^2
Wires 1 and 2 are made of the same metal. Wire 2 has twice the length and twice the diameter of wire $1 .$ What are the ratios (a) $\rho_{2} / \rho_{1}$ of the resistivities and (b) $R_{2} / R_{1}$ of the resistances of the two wires?
Watch the video solution with this free unlock.
EMAIL
PASSWORD