Resistance is proportional to length, and inversely proportional to cross-sectional area. Area depends on the square of the radius:  , so Student B's wire has  times the cross-sectional area of Student A's wire. In order to compensate for the increased area, Student B must make his wire  the length of Student A's wire. This can be shown mathematically using the equation for resistance:

Recall the formula for resistance  is given by

, where  is the cross sectional area,  is resistivity, and  is length. 

Solve for resistivity:

From this, we can tell that resistivity is proportional to cross sectional area by:

Since  is doubled, and resistance  and length  are constant, resistivity  will also be doubled. 

Resistivity  is given by:

, where  is the resistance,  is the area, and  is the length.

Since both have identical resistances and area, the first resistor will have half the resistivity since it has twice the length. Therefore the resistivity relation is

The equation for resistance is as follows: . Where  is resistivity,  is the length of the wire, and  is the cross section of the wire which can be found using .

The voltage drop from the source to the station (the "load") indicates that there is an internal resistance in the wire. According to the voltage law, the total amount of voltage drop is equal to the total amount of voltage supplied. Since  was supplied, and  drops at the station, that means that  drops along the wire. 

Now that the voltage drop across the wire is known, Ohm's law will give the resistance of the wire:

The resistivity of the wire is equal to the resistance per unit length, therefore, in order to find resistivity you divide the total resistance by the length:

The equation for resistance is as follows:

Where  is the resistance,  is the resistivity of the material,  is the length of the material and  is the cross-sectional area of the material. Looking at this equation, by doubling the area we effectively reduce the resistance by a factor of two.

There exist a formula that directly relates resistance and resistivity. The formula is . 

 is resistance,  is resistivity,  is length, and  is cross sectional area. Solving for , we get . Plugging in our givens, we get .

In order to find the power loss in R1, we need to know the current flowing through R1. Since it is not in parallel with anything, all of the current flowing through the circuit will flow through R1. To find the current flowing through the circuit, we will need to first find the total equivalent resistance of the circuit.

To do this, we first we need to condense R3 and R4. They are in series, so we can simply add them to get:

Now we can condense R2 and R34. They are in parallel, so we will use the following equation:

The equivalent circuit now looks like:

Since everything is in parallel, we can simply add everything up:

Now that we have the total resistance of the circuit, we can use Ohm's law to find the current:

Rearranging for current, we get:

Now that we know the current flowing through R1, we can use the following equation to find the power loss:

Since we don't know the voltage drop across R1 (although we can calculate it), we can substitute Ohm's law into the equation:

Plugging in our values, we get:

To calculate the power loss through R2 we need to either calculate the current flowing through it or the voltage drop across it. Calculating the current will be one less step, so we'll use that method.

We need to first calculate the total equivalent resistance of the circuit. Since the two resistors are in series, we can simply add their resistance values.

Then we can use Ohm's law to calculate the current through the circuit:

Next, use the expression for power:

Substituting Ohm's law for the voltage across the resistor, we get:

We know the voltage of the circuit, so we simply need the current through the circuit.

The new circuit has two resistors in parallel: R2 and the new one attached. To find the equivalent resistance of these two branches, we use the following expression:

In this new equivalent circuit, everything is in series, so we can simply add up the resistances:

Now we can use Ohm's law to calculate the total current through the circuit:

Now we can use the equation for power: