Question

a wire of resistance is 40 ohms is folded to double find the new resistance

Answer 1

Answer:

After folding the wire to double, the new resistance of the wire, R', calculated is $10\Omega$.

Explanation:

Data given,

The resistance of the wire of length L and area of cross-section A, R = $40\Omega$

After folding the wire to double, the length becomes half and the area becomes double.

Therefore, new length ( L' ) = $\frac{L}{2}$

And the new area ( A' ) = 2A

The new resistance of the wire, R' =?

Now, we know that,

• R = $\frac{\rho L}{A}$

Here, ρ = resistivity

After folding the wire to double, the new resistance:

• R' = $\frac{\rho L'}{A'}$
• R' = $\frac{\rho L}{2\times 2A}$
• R' = $\frac{\rho L}{4A}$

We know that, from the above equation: R = $\frac{\rho L}{A}$

Therefore,

• R' = $\frac{R}{4}$
• R' = $\frac{40}{4}$ = $10\Omega$

Hence, the new resistance of the wire, R' = $10\Omega$.