Question

A piece of wire of resistance $4\Omega$ is bent through ${180}^{\circ }$ at its mid point and the two halves are twisted together, then resistance is :

• A. $1\Omega$
• B. $2\Omega$
• C. $5\Omega$
• D. $8\Omega$

Answer 1

Answer: (A)

$1\Omega$

The resistance of a conductor and length of the conductor are directly proportional to each other. Likewise, the area of cross section of the conductor and the resistance as inversely proportional to each other. As the length of the conductor increases, the resistance of the conductor also increases and as the area of cross section increases, the resistance decreases. The relation between the length, area of cross section and resistance of a conductor is given as $R=\rho \frac{l}{A}$.
In this case, a piece of wire is bent through 180 degrees at its mid point and the two halves are twisted together. So, the length is halved (l/2) and the area of cross section is doubled (2A).
The equation is rewritten as
$R^{{}^{\prime }}=\rho \frac{\frac{l}{2}}{2A}=\frac{1}{4}\times \rho \frac{l}{A}$.
Solving both equations we get,
$R^{{}^{\prime }}=\frac{R}{4}$.
Therefore, the resistance becomes 4/4 = 1 ohm.