Question

Q1.Students are performing a lab using lengths of wire as resistors. The two students have wires made of the exact same material, but Student B has a wire that has twice the radius of Student A's wire. If Student B wants his wire to have the same resistance as Student A's wire, how should Student B's wire length compare to Student A's wire?

Answer 1

Explanation:

To find:

When will student A's and B's wire will have same resistance?

Calculation:

Let's consider A's wire to have radius r, so it's area is πr².

Now, B' s wire radius will be 2r, so area is π(2r)² = 4πr².

Now, we know that general equation for resistance is :

$\rm \: R = \dfrac{ \rho l}{ area}</p><p>$

Now, for student A :

$\rm \: R = \dfrac{ \rho l}{\pi {r}^{2} }$

Now, for student B (for same resistance):

$\rm \: R = \dfrac{ \rho l_{2} }{4\pi {r}^{2} }$

Now, dividing the Equations:

$\rm \implies \: 1 = \dfrac{ l_{2} }{4l}$

$\rm \implies \: l_{2} = 4l$

So, length of wire of B have to 4 times of the length of wire of A.

Answer 2

The wire length of student B should be 4 times as long as the wire length of student A.

Explanation:

Let the length of the resistor of student A be $l_{a}$

the area of the resistor of student A be $A_{a}$

the length of the resistor of student B be $l_{b}$

the area of the resistor of student B be $A_{b}$

Given, $r_{b}=2r_{a}\\\therefore A_{b}=4A_{a} [\because A=\pi r^{2}]$

We know, $R=\rho\frac{l}{A}$

For

$R_{a}=R_{b}\\\rho_{a} \frac{l_{a}}{A_{a}}=\rho_{b} \frac{l_{b}}{A_{b}}$

As materials are the same,$\rho_{a}=\rho_{b}$

$\therefore \frac{l_{a}}{A_{a}}=\frac{l_{b}}{A_{b}}\\\frac{l_{a}}{A_{a}}=\frac{l_{b}}{4A_{a}}\\\therefore l_{b}=4l_{a}$

The wire length of student B should be 4 times as long as wire length of student A.

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