The formula for resistance is: 

Above,  is length,  is the cross-sectional area of the wire, and  is the resistivity of the material, and is a property of that material. The resistivity is constant for a given material, and thus cannot be changed by altering the dimensions of the wire.

Let’s start with the resistivity equation

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

This can be simplified to 

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

Many things fallout which leaves us with

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

So we can substitute this into our equation

The length of the wire drops out of the equation

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

Take the square root of both sides

Therefore the ratio of the long to the short wire is

  Or

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

Length = 

Resistance = 

ρ of Tungsten = 

The equation for resistivity is

We can rearrange this equation to solve for 

In this case the area of the wire is the area of a circle which is equal to 

We can rearrange this to get the radius by itself

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

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 . 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

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

Therefore 

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,  would have the greatest resistance flowing through it.

To begin, let us start with the resistor and the  resistor that are in parallel. In parallel we can add resistors through the equation 

This new resistor is now in series with the two  resistors.  In series we can just add these resistors up.

This new resistor is now in parallel with the  resistor. In parallel we can add resistors through the equation 

This new resistor is now in series with the  and  resistor.  In series we can just add these resistors up.

Now that we have the equivalent resistance of the circuit.  We can now determine the current flowing out of the battery.

Rearrange to solve for current.

One of the best ways to work through a problem like this is to create a V= IR chart for all the components of the circuit.

We know that the current that is flowing out of the battery is the current that is flowing through both by 5 and 3 Ohm resistors since all of these are in series.  So we can put this information into our chart.

Using Ohm’s Law we can determine the voltage for each of these two resistors.

We can now use Ohm’s law and our voltage to determine the current going through the 14 Ohm resistor.

Next we can analyze the current in the Junction between the 5, 14 and 4 Ohm resistors.  Kirchoff’s laws state that the sum of the current flowing in and out of junction must equal 0.

We have  flowing in from the  resistor and  flowing out to go through the  resistor.

So the current going through the 4 Ohm resistor is 3 amps.  This will be the same for both 4 ohm resistors as both have the same current going into and out of the junctions near them.  We can then use this to determine the voltage drop across each of the resistors.

Now let’s add this to our chart.

We can now use Kirchoff’s loop law to determine the voltage across the 10 Ohm resistor.  Let’s analyze the loop that goes from the 5, to the 4 to the 10 back through the 4 and then through 3 Ohm resistor.

We can now use Ohm’s Law to determine the current through the 10 Ohm resistor.

To begin, let us start with the  resistor and the  resistor that are in parallel. In parallel we can add resistors through the equation 

This new resistor is now in series with the two  resistors.  In series we can just add these resistors up.

This new resistor is now in parallel with the  resistor. In parallel we can add resistors through the equation 

This new resistor is now in series with the  and  resistor.  In series we can just add these resistors up.

Now that we have the equivalent resistance of the circuit.  We can now determine the current flowing out of the battery.

Rearrange to solve for current.

One of the best ways to work through a problem like this is to create a V= IR chart for all the components of the circuit.

We know that the current that is flowing out of the battery is the current that is flowing through both by 5 and 3 Ohm resistors since all of these are in series.  So we can put this information into our chart.

Using Ohm’s Law we can determine the voltage for each of these two resistors.

We can now use Kirchoff’s loop law through the loop of the 5, 3, and 14 Ohm resistor to determine the voltage that is traveling through the 14 Ohm resistor.

We can now use Ohm’s law and our voltage to determine the current going through the 14 Ohm resistor.

We can now add this information to our chart.

Next we can analyze the current in the Junction between the 5, 14 and 4 Ohm resistors.  Kirchoff’s laws state that the sum of the current flowing in and out of the junction must equal 0.

We have 4A flowing in from the 5 Ohm resistor and 1A flowing out to go through the 14 Ohm resistor.

So the current going through the 4 Ohm resistor is 3 amps.  This will be the same for both 4 ohm resistors as both have the same current going into and out of the junctions near them.  We can then use this to determine the voltage drop across each of the resistors.

Now let’s add this to our chart.

We can now use Kirchoff’s loop law to determine the voltage across the 10 Ohm resistor.  Let’s analyze the loop that goes from the 5, to the 4 to the 10 back through the 4 and then through 3 Ohm resistor.

We can now use Ohm’s Law to determine the current through the 10 Ohm resistor.

We can now add this information to our chart.

We can now analyze the junction between the 4, 10 and 15 Ohm resistor.

Kirchoff’s laws state that the sum of the current flowing in and out of the junction must equal 0.

We have 2A flowing in from the 4 Ohm resistor and 1.2A flowing out to go through the 10 Ohm resistor.

To begin we need to simplify the circuit to get the equivalent resistance.  Let’s start with the 3 and 6 Ohm resistors in parallel.

Now we can add this resistor to the 4 Ohm resistor as they are in series.

We can now determine the current coming out of the battery using Ohm’s Law.

Rearrange to solve for current.

The current coming out of the battery will be the same current that moves through the 4 Ohm resistor.  So we can determine the voltage drop across the 4 ohm resistor.

We can then use Kirchoff’s loop law to determine the voltage drop across the 6 Ohm resistor.  Let’s analyze the loop that goes from the battery to the 4 ohm resistor and through the 6 ohm resistor.