Let’s start with the resistivity equation

\(\displaystyle R = \rho \frac{L}{A}\)

The area of a wire is the area of a circle.  So let’s substitute that into our equation

\(\displaystyle A = \pi r^2\)

\(\displaystyle A = \pi \frac{d^2}{4}\)

\(\displaystyle R = \rho \frac{L}{\pi \frac{d^2}{4}}\)

This can be simplified to 

\(\displaystyle R = 4\rho \frac{L}{\pi d^2}\)

Since we know that both resistors have the same resistivity and the same resistance, we can set these equations equal to each other.

\(\displaystyle 4\rho \frac{L_1}{\pi d_1^2} = 4\rho \frac{L_2}{\pi d_2^2}\)

Many things fallout which leaves us with

\(\displaystyle \frac{L_1}{d_1^2} = \frac{L_2}{d_2^2}\)

We know that the second wire is twice the length as the first

\(\displaystyle L_2 = 2L_1\)

So we can substitute this into our equation

\(\displaystyle \frac{L_1}{d_1^2} = \frac{2L_1}{d_2^2}\)

The length of the wire drops out of the equation

\(\displaystyle \frac{1}{d_1^2} = \frac{2}{d_2^2}\)

Now we can solve for the diameter of the longer wire.

\(\displaystyle d_2^2 = 2d_1^2\)

Take the square root of both sides

\(\displaystyle d_2 = \sqrt{2} d_1\)

Therefore the ratio of the long to the short wire is

\(\displaystyle \frac{\sqrt{2} d_1}{d_1}\)

  Or

\(\displaystyle \sqrt{2}\)

We will use the resistivity equation to solve for this.  We know

Length = \(\displaystyle 1.0m\)

Resistance = \(\displaystyle 0.32 \: ohms\)

ρ of Tungsten = \(\displaystyle 5.6x10^-8 \Omega m\)

The equation for resistivity is

\(\displaystyle R = \rho \frac{L}{A}\)

We can rearrange this equation to solve for \(\displaystyle A\)

\(\displaystyle A =\rho \frac{R}{L}\)

In this case the area of the wire is the area of a circle which is equal to \(\displaystyle \pi r^2\)

We can rearrange this to get the radius by itself

\(\displaystyle r = \sqrt{\frac{\rho R}{L \pi}}\)

\(\displaystyle r = \sqrt{\frac{(5.6x10^-8)(0.32)}{(1)(\pi)}}\)

\(\displaystyle r = 7.5x10^-5m\)

To find the diameter we need to multiply this value by 2.

\(\displaystyle d = 1.5x10^-4m\)

At higher temperature, the atoms are moving more rapidly and are arranged in a less orderly way.  Therefore it is expected that these fast moving atoms are more likely to interfere with the flow of electrons.  If there is more interference in the flow of electrons, then there is a higher resistivity.

Resistors are in parallel when the electric current passes through two or more branches or connected parts at the same time before it combines again. After the current leaves the battery it travels to \(\displaystyle R_{2}\). Then the current splits and travels between the other two resistors before the current coming back together and connecting back to the battery.

Adding resistors in parallel decrease the overall equivalent resistance as they are added using the equation

\(\displaystyle \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}+\cdots\)

Since the overall equivalent resistance is decreased, if the voltage is constant the overall current would increase.

\(\displaystyle V = IR\)

Therefore 

\(\displaystyle I = \frac{V}{R}\)

This shows the inverse relationship between the two values.

As we examine the circuit we first come across the first two resistors. These resistors are in parallel. In parallel circuits, the current splits to go down each branch. In this case, the resistors are of equal value meaning that the current is split evenly between the two.

We then come across another parallel branch with two resistors on one side and a single resistor on the other side. The two resistors in series add up to create more resistance on the top branch. The single resistor on its own has less resistance.

Since current always chooses the path of least resistance, more current will flow through the single resistor, than through the branch with two resistors in series.

Therefore, \(\displaystyle R_5\) would have the greatest resistance flowing through it.