Question

1. a wire of resistivity \\( \

ho \\) must be replaced in a circuit by a wire of the same material but 4 times as long. if, however, the resistance of the new wire is to be the same as the resistance of the original wire, the diameter of the new wire must be
a) the same as the diameter of the original wire.
b) \\( 1 / 2 \\) the diameter of the original wire.
c) \\( 1 / 4 \\) the diameter of the original wire.
d) 2 times the diameter of the original wire.
e) 4 times the diameter of the original wire.

Explanation:

Step1: Recall the resistance formula

The resistance formula is \(R =
ho\frac{l}{A}\), where \(A=\pi(\frac{d}{2})^2\) ( \(d\) is the diameter), so \(R =
ho\frac{4l}{\pi d^{2}}\).
Let the original length be \(l_1\), diameter be \(d_1\), and the new length be \(l_2 = 4l_1\), diameter be \(d_2\). Since \(R_1=R_2\), \(
ho\frac{4l_1}{\pi d_{1}^{2}}=
ho\frac{4l_2}{\pi d_{2}^{2}}\).

Step2: Substitute \(l_2 = 4l_1\) into the equation

Substitute \(l_2 = 4l_1\) into \(
ho\frac{4l_1}{\pi d_{1}^{2}}=
ho\frac{4l_2}{\pi d_{2}^{2}}\), we get \(\frac{l_1}{d_{1}^{2}}=\frac{4l_1}{d_{2}^{2}}\).
Cross - multiply: \(d_{2}^{2}=4d_{1}^{2}\).
Take the square root of both sides: \(d_2 = 2d_1\).

Answer:

D. 2 times the diameter of the original wire.