Question

The resistance of a conductor depends on which factors and how? The resistance of a wire is \( R \). What will be the resistance of the wire when the wire is of double length and half the radius of the same material?

Hint:

Resistance increases with increase in length and decreases with increase in thickness (area). Use \( R \propto \frac{L}{r^2} \) for quick estimation.

Answer 1

Step 1: Factors affecting resistance.
The resistance \( R \) of a wire depends on: \[ R = \rho \frac{L}{A} \] where, \( \rho \) = resistivity of material (depends on the substance),
\( L \) = length of the wire,
\( A \) = cross-sectional area of the wire.
Step 2: Dependence on dimensions.
- Resistance is directly proportional to the length of the wire.
- Resistance is inversely proportional to the area of cross-section of the wire.
Step 3: Given condition.
Original resistance = \( R = \rho \frac{L}{A} \).
New wire has: Length \( L' = 2L \), and radius \( r' = \frac{r}{2} \).
Therefore, area of cross-section \( A' = \pi (r')^2 = \pi \left(\frac{r}{2}\right)^2 = \frac{A}{4} \).
Step 4: Calculate new resistance.
\[ R' = \rho \frac{L'}{A'} = \rho \frac{2L}{A/4} = \rho \frac{2L \times 4}{A} = 8 \rho \frac{L}{A} = 8R \] Step 5: Conclusion.
Hence, when the length is doubled and radius is halved, the new resistance becomes \[ \boxed{R' = 8R} \]