Question

A piece of wire, having resistance R, is stretched thrice to its original length.
(a) What will be its new resistance?
(b) It is then cut in a ratio of 1:2 and connected in parallel. Find the new resistance combination.

Answer 1

Hint: We need to understand the relation between the resistance of a wire, its length, and the type of network combination to solve this problem. We can use our ideas in these two situations above to get the solution for the new resistance.

Complete step-by-step solution
We know that the resistance of a wire is dependent on the length of the conducting wire, the area of the cross-section of the wire, and the resistivity of the material of the wire. The resistivity of a material is a characteristic of the material which is independent of the physical dimensions other than the temperature. It is given as –
\[R=\rho \dfrac{l}{A}\]
Let us consider at a given temperature the two situations are employed. The resistivity will be constant and the resistance will be dependent on the length and area of the wire.
(a)The wire is stretched three times as its original length. The increase in length to such a scale has an effect on the area of the wire. We know that the volume of the wire remains constant, so we can find the new area for the new length as –
\[\begin{align}
  & V=Al \\
 & \Rightarrow V=\pi {{r}^{2}}l \\
 & \text{The volume remains same, but the length is tripled}\text{.} \\
 & \Rightarrow V=A'(3l) \\
 & \Rightarrow \pi {{r}^{2}}l=\pi r{{'}^{2}}(3l) \\
 & \Rightarrow r'=\dfrac{r}{\sqrt{3}} \\
 & \Rightarrow A'=\pi r{{'}^{2}} \\
 & \therefore A'=\dfrac{A}{3} \\
\end{align}\]
Now, we can find the new resistance for this system as –
\[\begin{align}
  & R'\propto \dfrac{l'}{A'} \\
 & \Rightarrow R'\propto \dfrac{3l}{\dfrac{A}{3}} \\
 & \therefore R'=9R \\
\end{align}\]
The new resistance is nine times the initial resistance of the wire.
(b) Now, the same wire with the new length and new area is cut in the ratio 2:1. The area will be the same for both the pieces. The length of the two wires will become –
\[\begin{align}
  & {{l}_{1}}=2l \\
 & {{l}_{2}}=l \\
\end{align}\]
We can find the resistance for each of the wires as –
\[\begin{align}
  & {{R}_{1}}\propto \dfrac{{{l}_{1}}}{A'} \\
 & \Rightarrow {{R}_{1}}\propto \dfrac{2l}{\dfrac{A}{3}} \\
 & \therefore {{R}_{1}}=6R \\
\end{align}\]
For the second wire,
\[\begin{align}
  & {{R}_{2}}\propto \dfrac{{{l}_{2}}}{A'} \\
 & \Rightarrow {{R}_{2}}\propto \dfrac{l}{\dfrac{A}{3}} \\
 & \therefore {{R}_{2}}=3R \\
\end{align}\]
Now, the two wires of resistance ‘6R’ and ‘3R’ are connected in parallel. We can find the equivalent resistance of the system as –
\[\begin{align}
  & \dfrac{1}{R''}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}} \\
 & \Rightarrow \dfrac{1}{R''}=\dfrac{1}{6R}+\dfrac{1}{3R} \\
 & \Rightarrow R''=\dfrac{18{{R}^{2}}}{9R} \\
 & \therefore R''=2R \\
\end{align}\]
The new resistance in parallel combination will be twice the initial resistance.
So, the resistance of the wire in the cases become –
(a) 9R
(b) 2R
This is the required solution.

Note: The resistance of the wire is dependent on the length, the cross-sectional area, and the resistivity of the material of the wire. We also understand that the same wire when cut and connected in parallel can give entirely different resistance as we have seen.